Quants

quants concepts

Compiled by Amul Badjatya From pagal guy threads=====================================================================32^32=(1024)^ 16=(24)^16= (576)^8=(24)^8=(24)^4=24^2=576 .Hence second last digit=7. ========================= An extension of the same problem: 1111.......n times is a composite number for sure only when n is composite number. ================================== oday's funda...it's a combination of few of my old posts... factorial based questions asking no. of zeroes and max power of sum integer. Find the no. of zeroes at the right end of 300! for every zero, we require 10..n every 10 is made up of 5x2. in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300! 300/5 = 60 (because every fifth no. is a multiple of 5) 300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12 300/125 = 3 (because multiples of 125 have three 5s in it) or, 12/5 = 2 now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74. we might also get the same type of questions in a different form, 500! is divisible by 1000^n...what is the max. integral value of n? now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3 500/5 = 100 100/5 = 20 20/5 = 4 100 + 20 + 4 =124 124/3 = 41.33 max integral value is 41. 500! is divisible by 99^n...what is the max. integral value of n? now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer

500/11 = 45 45/11 = 4 ans will be 49. so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example: 100! is divisible by 160^n...what is the max. integral value of n? now 160 = 2^5 * 5^1. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5. 100/2 = 50 50/2 = 25 25/2 =12 12/2 = 6 6 /2 = 3 3/2 = 1 add 'em all...97. 97/5 = 19. so the answer wud be 19 had v taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer... when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer. 50! is divisible by 144^n...what is the max. integral value of n? 144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor... count 2s 50/2=25 25/2=12 12/2=6 6/2=3 3/2=1 sum=47 answer = 47/4 = 11. count 3s 50/3=17 17/3=5

5/3=1 sum = 23 23/2 = 11 a tie...else the smaller value wud have been the answer. 300! is divisible by (24!)^n. what is the max. possible integral value of n? such questions are tricky...when u expand 24!...u get 1x2x3...24. in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x... when v expand 300!...v get a 23 in 23, 46,69,92.... total no of multiples of 23 in 300! will be 300/23 = 13, forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert. 256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end? such questions are same as finding maximum power of 576 in 256! 576 = 2^6 x 3^2 to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain. total 2s in 256! = 255 hence, no. of zeroes = 256/6 = 42. just to check...3s = 126, 126/2 = 63>42 ans-42

Questions based on this concept 400! is divisible by x^n. what is the max. possible integral value of n if the value of x is: Q1. Q2. Q3. Q4. Q5. Q6. Q7.

300 99 500 320 770 5200 270

Q8. 686 Q9. 338 Q10. 13000 Answers... 49, 39, 33, 66, 39, 32, 65, 22, 16, 32 (the answer is not 33, this one is actually tricky! ) 200! is divisible by (x!)^n...whats the max. possible value of n when x = Q11. 25 12. 35 13. 50 14. 100 15. 70 16. 300 17.15 answers... 8, 6, 4, 2, 3, 0, 16 300! is expanded and expressed in base x. find the no. of zeroes at the right end of this expression when x= 18. 19. 20. 21. 22.

25 15 35 39 98

answers...37, 74, 48, 23, 24 do lemme know if there's any problem at all. cheers! maxximus ============================ Today's concept: Finding out smallest no. which leaves specific remainders with specific divisors. Type # 1. find smallest no. other than k, that leaves remainder k when divided by w,x,y... to solve such questions, take lcm of w,x,y...and add k to it. e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9... take lcm of 6,7,8,9 and add 4 i.e. 504 + 4 = 508 Type # 2 find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively. unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23 Type # 3 Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11, we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208 now we have one more condition...remainder 1 with 11. concept => to a no. if v add lcm of divisors...the corresponding remainders dont change. i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled. so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1 208%11 = 10 210k%11 = k therefore, 10 + k shud leave remainder 1 when divided by 11. hence, k = 2. and the no. is 208 + 210 x 2 = 628 e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7 for first 3 conditions....no. is 120 + 2 = 122 hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362 Type # 4 What if there is no relation between divisors n remainders? e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13. we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant. in such cases...take 1 case n target another case... e.g. i take the case 7 with 13...and target 6 with 11. which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6. a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72 now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7. to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied... hence the no. is of the form 72 + 143 k. 72 + 143k % 7 = 2 + 3k now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7.. a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3. hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501. now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501. For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically. there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same... few points to be noted *you can always re-check ur answer **at times, u can use options to solve such questions. ***dont let concepts go away believing such questions can be easily dealt with thru options...the question may not always be find the smallest no. which...... at times it may be ..find the sum of integers of smallest no. which leaves remainders...blah blah... ****there may be a case when they put an option which satisfies all the conditions but is not the smallest poss value...n put another option...our favorite...none of these!!! lets not undermine genius of cat makers!! questions for practice... find the smallest no. that leaves remainder (s) Q1. 2 when divided by 3,5,6 or 9 (other than 2) Q2. 2,5,7 when divided by 7,10 and 12 respectively Q3. 1,2,3,4 with 3,4,5,7 respectively.

Q4. Q5. Q6. Q7.

6 3 2 1

with with with with

7,8,9,10 and 3 with 11. 6, 0 with 11, 3 with 5, 7 with 8 5, 7 with 8, 3 with 4, 5 with 7&11 11, 4 with 5, 9 with 10, 7 with 9.

hope the post helps...puys...come up with ur own...sweet methods...which are confined in ur sharp brains...i'll also appreciate if u can come up with feedbacks/suggestions...like today i thought its better to give answers a day late... regards maxximus ============================== Hi Puys! Here is something i used and doubt if many are aware of... This is a method to divide quickly (very Quickly) and get results with amazing accuracy... But let me forewarn you, it needs good amount of practice before you can even think of trying it in exam hall... And yes, this is not my creation. Credits to Rahul and company who shared this with me 2 yrs ago so here we Go!!! The approach would be to get the denominator to either a 100 or a 1000 because that is what percentages is all bout. Simply focus on the fact that how do the given denominators reach 100/1000. I have left some blank because they are very obvious. The following are the stations between 100 and 1000 100 111 125 133 150 166 182 200 222 250 273 300 333 364 400 455 500 555 600 666 700 750 800 833 875

-

Reduce 10% Multiply by 8 Reduce 1/4 Reduce 1/3 Multiply by 6 Add 10%

- Reduce 10% - Add 10% - Reduce 10% - Add 10% - Reduce 10% - Add Half - Add 1/3 - Add 20% - Add 1/7

900 910 - Add 9% 950 - Add 5% 1000 So Task (1) you have to mug up the values of these stations. It is very important that these values are memorized because this will help you in knowing which number to reach from any given number. e.g. if the denominator is 887 you know you have to reach for 900 or 875 and so on. Task (2) Practice! Below is an approach to tackle (three digit /three digit) with consummate ease. We shall attempt to understand it with examples. Example 1: What is 145/182 ------------------79.5 Steps 1. add 10% to numerator and denominator... 2. it becomes 159/200 ...............which is 79.5 ( the answer from the calculator is 79.6) Example 2: 123/178....??? Step1:- which station is closest to it?.............................200?...or some say 150. Either is good. (identification took 2 seconds) Step 2:- what do I have to do to go from 178 to 200/150..................add 22/subtract 28... (another 2 seconds.) Step 3:- so if I add 22................i am actually adding slightly less than 13% of 178 to itself. This part is tricky.. here is how I got it: This is how u Our number is 22 definitely if 10% = 17.8

need 178. more then

to think--Thus, 10% of it is 17.8 than 10% . 5% = 8.xx

And 2.5% = 4.xxx (don't even bother to calculate xxx) which means its around 12.5........or 13 or 12.......... Add same to numerator...........10% of 123 = 12.3 1% = 1.23 so 2% = 2.46 So 12% = 15? So it becomes 138/200 = 69%? (calculator answer is 69.1). All you require to calculate is what is 10% 1%...and approx stuff...any damn calculation works in less than 8 seconds.:satisfie: Some thing like (2456*4567 - 2134*3214)/2134*3214 will taken 10 seconds maximum... it works coz I had that time 1.5 years ago..and I aint kidding..

Lets take one more. Example 3: 532/745? Tough??? This's how it can be approached: # Nearest station.........750.................so add 5..............about 1% or less..add 1% damnit # Numerator now is 537.............(added 1%) # Fraction is 537/750 add 1/3 each...........it is 71 something. Remember here don't even attempt to do...537/3. Because denominator is a 1000 and not a 100. So one digit is redundant. So all I do is........53/3 approx..18...plus..53 = 71%..(will do if answer are spaced...) answers not spaced? Then 537 +537/3 = 537+179 = 716 which makes it 71.6 (calculator is 71.5) Dont worry if the last few statements were difficult to digest. Try solving a few questions and you'll get the crux of it. Remember in CAT we don't find answers... we choose them! Happy Computing Libocta ---------------------------------------------------------------------Once an IIMCian, an IIMCian for lifetime! ================================================================= @ cat 2007 ...cool question bro...wud bring in a new concept...thats wat we're look'n for...to let our brains work in unexplored areas...this question helps us c another aspect while approaching a question... here's the solution...or lets say approach... 4/3x22/7xR^3 + 4/3x22/7x r^3 is the sum of volumes...the terms in denominator shud be cancelled with R^3 as well as r^3. but terms in numerator cannot be cancelled since radii are integers... hence each term is divisible by 22x4 i.e. 88 and so is the sum... as all the options are divisible by 8 (since last 3 terms of each value is divisible by we just need to check which option is divisible by 11...thats option 2. so that must be the correct option.... am i right cat 2007? do keep pouring in.... @ libocta...great to have u around...ur expertise wud be very helpful!! with warm regards maxximus ============================================================

Fisrt of all lets find out whats the cube of all single digit numbers

1 cube = 1 2 cube = 8 3 cube = 27 4 cube = 64 5 cube = 125 6 cube = 216 7 cube = 343 8 cube = 512 9 cube = 729 now, from the question we see that last digit of the number 328509 is 9 so the number whose cube is 328509 should end with 9 (since only cube of 9 ends with 9) also, 50 cube = 125000 60 cube = 216000 70 cube = 343000 our number lies between 60 cube and 70 cube, so the 1st digit has to be 6, and we already have the last digit as 9 so the number is 69.

++++++++++++++++++++++++++========================================

a^n - b^n is always divisible by a-b a^n - b^n is divisible by a+b when n is even. a^n + b ^n is divisible by a+b when n is odd. a^n + b^n + c^n +... is divisible by a+b+c+.... when n is odd. when last n digits of a no. are divided by 2^n, the remainder is same as the remainder when the entire no. is divided by 2^n.

Answers to yesterday's questions. 1. smallest no. that must be added to 1000 so that the resultant no. leaves remainders 2,3,4,5 with 5,6,7,11 respectivelt----667 2. smallest no. that leaves remainders 1,2,5,6, when divided successively by 2,3,4,23.----such a no. doesnt exist as 5 rem with aint possible. 3. smallest no. that leaves remainders 4 everytime when successively divided by 7,5,10,13 respectively.----872 4. a no. leaves remainders 2,5,3,7 when successively divided by 3,7,6,9. what is the remainder when this no. is divided by 126?----80 5.find the largest no. that leaves same remainder when it divides 2345,7645,9845,6595 and 10095.----50

6.a no. when divided 88 leaves remainder 3. what is the remainder when its divided by 11? ----3 7.a no. when divided by 391 leaves rem. of 49. find the remainder when its divided by 39...options 29,10,none of these,cannot be determined.---cannot be detrmnd kudos to rockeezee who got most of them right!

questions for today... 1. 32^23 + 17^23 is definetly divisible by.... a. 49 b. 15 c. 49 & 15 d. none of these. 2. 32^23 - 17^23 is definetly divisible by.... a. 49 b. 15 c. 49 & 15 d. none of these. 3. 32^232 + 17^232 is definetly divisible by.... a. 49 b. 15 c. 49 & 15 d. none of these. 4. 32^232 - 17^232 is definetly divisible by.... a. 49 b. 15 c. 49 & 15 d. none of these. 5. 32^2321 + 17^2321 + 13^2321 + 18^2321 is definetly divisible by.... a. 40 b. 20 c. 80 d. all of these. e. none of these. what is the remainder when 42527152653425416242624272427215287 is divided by : 6. 16 7. 32 8. 64

regards maxximus =================================

Today's concept...to solve tricky questions based on no. of factors of a number. Despite the interest shown in the concept yesterday, not many cud solve the question efficiently...so, best concept for the day can be to discuss yesterday's problems...have a quick, practical solution to them...and practice a lot of similar, different questions on same concept. here we go... Find the smallest no. that has exactly..... 1. 16 factors 2. 12 factors 3. 60 factors questions are lil tricky...but if u get the concept...they become child's play...see how... 16 factors...that mean product of (powers +1) of all the prime nos = 16. now 16 can be achieved in following ways... 16 8x2 4x4 2x4x2 2x2x2x2. by sheer common sense, v can say the highest power shud go to the smallest prime no. i.e. 2....and as v proceed, smaller powers shud be given to higher prime nos. Concept: powers shud reduce and the corresponding prime nos. shud increase. so... 16...the no. is 2^15 8x2...the no. is 2^7 x 3 4x4...the no. is 2^3 x 3^3. 2x4x2...the no. is 2^3 x 3 x 5 2x2x2x2.....the no. is 2x3x5x7. first three nos can be easily discarded as they are too big...just calculate last two nos, they are 120 and 210...120 is smaller and hence the answer. important: please avoid craming...in few cases last way might give best answer..in other cases, 2nd last one...its always advicable to form patterns n check the closer ones. 2. 12 factors

12 = 12 or 4x3 or 2x2x3. easily answer wud be 2^3 x 3^2 or 2^2 x 3 x 5 the corresponding values are 72 n 60. hence, the answer is 60. similarly, Q3 also. 4. how many factors of 27000 are perfect cubes? i realy wonder y nobody cud get this right...u just need to form combinations and check which combinations give u cubes...they are 27000,27,1000, etc. am not discussing this question...i hope when i give a similar question today...i get few correct answers. This question is still open for answers/discussion (so are others...but if u can answer this...with an xplanation, it'll be gr8 ) how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast... 5. 3 terms 6. 4 terms 7. 5 terms 8. 6 terms

concept: to have first n last term as the given terms...the difference b/w the two terms shud be divisible by the common difference. so u basically have to find how many such common differences exist...for every common difference...u'll get a new AP. for practice, lets take a small interval [1,15] the difference is 15-1=14. now 14 is divisible by 1,2,7,14...four different integers... so v can have four different APs...if v take a common difference other than these four values...the last term wont lie in that AP. e.g if v take the common diff = 4, the AP wud be 1,5,9,13,17...see 15 didnt lie... so, 4 APs are possible. now if a conditions is attached...there shud be atleast three terms...it means that AP with 2 terms shud be neglected... 1,15 is an AP with jus 2 terms...so it shud be beglected...remember, an AP with 2 terms always lies in any interval. so the answer wud be 4-1 = 3 if the condition is atleast four terms...then the AP with 2 terms as well as the AP with 3 terms shud be neglected.

we know that an AP with 2 terms is bound to exist...lets c if an AP with 3 terms also exists. an AP with 3 terms will look like... 1, x, 15 see...there are 2 intervals... x-1 and 15-x. hence for a 3 term AP to exist, the difference shud be divisible by 2. since 14 is divisible by 2, we further reduce the answer by 1...so APs with atleast 4 terms are 3-1 =2 now, if the conditions is...atleast 5 terms...v need to check if AP with 4 terms exists... such an AP wud luk like... 1, x, y, 15 see, there are 3 intervals...since 14 is not divisible by 3, such an AP does not exist. so the answer remains 2. similarly, for atleast 6 terms, v check if 14%4 = 0...since no, the answer is again 2 for 7 terms 14%5 is not 0, the answer is 2 for atleast 8 terms...14%6 is not 0 so the answer is again 2. for 9 terms, 14%7 = 0. hence answer becomes 2-1 =1 keep on proceeding like this...the soln wont be so bulky...its been done like this for ease of understanding...for ease of calculation...see how to proceed... how many A.Ps are possible such that first term is 1235 and the last term is 3535 and there are atleast... 5. 3 terms 6. 4 terms 7. 5 terms 8. 6 terms 3535 - 1235 = 2300. 2300 = 23x2^2x5^2 . hence, no. of factors = 3x3x2 = 18. (check yesterday's concept if missed) how many APs...18. how many with atleast 3 terms? since 2300%1 = 0, 18-1 = 17 how many with atleast 4 terms? 2300%2 = 0, hence 17-1 = 16

how many with atleast 5 terms? 2300%3 =/ 0, answer remains 16. where =/ means not equal to how many with atleast 6 terms? 2300%4 = 0, 16-1 = 15 how many with atleast 7 terms? 2300%5 = 0, 15-1 = 14 how many with atleast 8 terms? 2300%6 =/ 0 , answer remains 14 how many with atleast 9 terms? 2300%7 =/ 0, answer remains 14 and so on... to check for atleast n terms, v need divisibility till n-2...i'll appreciate if u dont cram...but understand it...i've neva learnt it...its jus an observation... so, the answers to above 3 questions wud be 17,16,16,15.

finally... How many values of a are possible if x^2 + ax + 2400 has... 9. integral roots 10. roots which are natural nos. this is an actually tricky problem...sad that nobody came up with this...see... concept: to solve ax2 + bx + c =0, we break it as ax2 + mx + nx + c = 0, such that m*n = a*c. here, a*c = 1*2400 = 2400. so v need to find in how many ways can 2400 be expressed as product of 2 nos. every such pair of nos. will give a new value of the coefficient of x. 2400 = 2^5 x 5^2 x 3 no. of factors = 6x3x2 = 36. but these factors have to exist in pairs...e.g when v use one factor 2 (to express 2400 as 2x1200) the other factor...1200 is automatically used...so total pairs possible are ... 36/2 = 18. but relax...this is not it....again...equal no. of negative pairs exist...i.e. 2x1200 corresponds to -2 x -1200. although the product is same as the reqd product

i.e 2400...the sum is different...its 1202 n -1202...n v need to find different values of sum...hence the answer wud 18x2 = 36 again... 9. integral roots 10. roots which are natural nos. 9. for integral roots...the answer wud be 36 10. values wud be 1/2 the total possible values as negative roots aint allowed...36/2 = 18 is the answer Kudos to rockeezee...who got 5 correct answers...so too to junoonmba,vani for their active participation!!! wud love to see all of u gett'n of the following correct today... Today's questions... smallest no. that has exactly. 1. 20 factors 2. 36 gactors 3. 30 factors how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast... 4. 5 terms 5. 7 terms 6. 10 terms. how many of the factors of 640000 have 7. perfect square roots. 8. perfect fourth roots 9. perfect cube roots how many different values can 'A' take if x2 + Ax + 2500 has ... 10. integral roots 11. negative roots. 12. non-negative integral roots. 13. find the no. of factors 15! has (here comes the season of fresh concepts...which'll derive from older ones...he he) 14. find the no. of factors 18! has

Minimum value that A can take if x2 + Ax + 900 has... 15. Integral roots. 16. Negative integral roots. 17. Positive integral roots.

this is a long post...with several concepts and a wide variety of questions...i believe it'll take some time to grasp, solve n discuss the concept properly...so i wont post a new concept tomorrow...i've got messages suggesting i shud be a bit slow....so lets discuss this very concept till v master it...start solving guys...at any point of difficulty...i'll be there... Happy solving!!!!! with due regards maxximus

======================================================================== What is the smallest number which on multiplying with 12.5464646464646�. would make the product an integer???? a) 198 b) 1188 c) 2475 d) 12375 There is a concept behind solving such questions.. Would anyone like to try it out??? (Mostly for Newbies) Season 2007-2008: Results awaited: IIMs XLRI : Selected MhCET : 99.83 (146/200 ============================== Hie...sorry guys...was out whole day...so a late reply... before anything else, i shud give the answers... smallest no. that has exactly. 1. 20 factors---240 2. 36 factors----1260 3. 30 factors----720

how many APs are possible such that the first term is 2454 , last term is 4254 and there are atleast... 4. 5 terms----33 5. 7 terms----31 6. 10 terms.----29 how many of the factors of 640000 have 7. perfect square roots.----18 great work gk1, rockeezee...here's a more practical approach...640000=5^4x2^10 express this as 25^2 x 4^5. now since 25 as well as 4 are perfect squares...combination of their products will also be pefect squares...no. of combinations here wud be (2+1) x (5+1) = 18 8. perfect fourth roots----6 express it as 16^2 x 625^1 x 4...since 16, 625 have perf fourth roots...so will their cominations...i.e. 3x2 = 6 of them. 9. perfect cube roots----8 express it as 8^3 x 125^1 x 10...since 8,125 are perfect cubes...their combinations wud yield perf cubes...their no. are... (3+1) x (1+1) = 8 how many different values can 'A' take if x2 + Ax + 2500 has ... 10. integral roots----16 2500 = 2^2 x 5^4. hence no. of factors are 3x5 = 15. no. of pairs = {15 / 2 } = 8. {} = least integer function...plz check...rounded it off to higher value...the value is odd because of 50 x 50 = 2500...giving only one combiantion seen both wat...while other combinations give 2 different cominations when the order is changed...for example...250 x 10 and 10 x 250... but 50 + 50 will give a new sum 100...so it shud be counted. so the answer in above case wud be {15/2}x2 = 8x 2 =16. for better understanding...lets take one more example with smaller value of a x c. e.g. x2 + Ax + 4 = 0. A can +/- (2+2 or 4 +1)...so 4 combinations...while 4 = 2^2...hence having 3 factors...to get the answer, we'll perform the following opeartn...{3/2} x 2(for negative counterparts) = 4. 11. negative roots----infinitely many great job rockeezee...its not mentioned negative integral roots...so infinitely many... i know many of u tuk out answer bcoz of faith in my typing errors...n must have thought that i forgot to type integer...well, in that case..the answer is 16/2 = 8. as 0 cannot be a root of above expression.

12. non-negative integral roots----8 13. find the no. of factors 15! has ---- 12x7x4x3x2x2 14. find the no. of factors 18! has---- 17x9x4x3x2x2x2 ...great job rock, junoon...try coming up xplanations as well...18! has 16 2s, 8 3s, 3 5s, 2 7s, 1 11, 1 13, 1 17. added one to the no. of times each prime no. existed and multiplied them... Minimum value that A can take if x2 + Ax + 900 has---15. Integral roots---- (-901) while taking -900 x -1 = 900 16. Negative integral roots---- 60...thru 30 x 30 17. Positive integral roots--- -901 thru -900 x -1 kudos to rockeezee,gk1,junoonmba,iyervani!!! who got many right..... hope this post clears few concepts....wud suggest u all to kindly go thru this post n the original post once more to cement the concept... wud love to hear from u guys to make this thread more effective...as per the feedbacks so far, am trying to make it slow n steady... with due regards maxximus =========================== Today's concept....finding squares and close multiplications quickly... In a hurry guys....so few small concepts that'll help u save sum time...n avoid cramming....which i've always maintained is the best ill-preparation for cat. #1 finding squares. step 1. think of a base which is a multiple of 10 or 100 (whichever nearer to the no. whose square is to be determined.) suppose 34^2. a gud base wud be 30. express 34 as 30 + 4 (i.e. base + 4)...i'll call this base + deviation step 2. ldigit(s) before 0 in base x (no. + deviation) | deviation^2 where | is an imaginary line separating 2 parts of calculation. i.e. 3x(34 + 4) | 4^2 i.e 114 | 16. now the important thing....no. of digits on right side of the imaginary line shud be exactly same as the no. of zeros in our base..dont forget this!!!!

since our base 30 has only one 0 at the end, v can have only one digit on right side of the imaginary line....the less powerful digit...thats 6. and thus, one has to be carried to the left side....so 114 will become 115. so the expression becomes...115 | 6 hence, the answer is 1156. look at few more examples for practice... 28^2 base 20 2x(28 + 8 ) | 8x8 = 72 | 64 = 784 106^2 106 + 6 | 6x6 base 100 =112 | 36 =11236 . since 100 has 2 zeroes, there shud be 2 digits on right side of the line.... 103^2 base 100 103 + 3 | 3x3 =106 | 9 = 10609. ensure the no. of digits are exactly same as the no. of zeroes in base....so expressed 9 as 09....a 2 digit no. 312^2 base 300 3x(312 + 12) | 12^2 3x324 | 144 972 | 144 = 97344....1 of 144 carried.

smart ways of using this method....

98^2. now if v take base 100, v can avoid multiplication by 9. so base 100. 98 - 2 | -2^2 = 96 | 04 = 9604....funny ha?

197^2 base 200 2x(197 - 3) | -3^2 =388 | 09 = 38809. #2 : finding product of nos when they lie nearby. all digits before 0 in base x (first no. + deviation of second no.) | deviation of first no. x deviation of second no. rest, everything is same... examples... 27 x 22 = base 20 2x(27 + 2) | 7x2 = 58 | 14 = 594 103 x 108 base 100 = 103 + 8 | 3x8 = 111 | 24

=11124 97 x 102 base 200 97 + 2 | -3 x 2 =99 | -6 = 9900 - 6 = 9894

197 x 199 base 200 = 2x(197 - 1) | -3 x -1 = 392 | 03 = 39203 37 x 31 base 30 37 + 1) x 3 | 7x1 = 114 | 7 = 1147. 32 x 24 base 30 32 - 8 | 2 x -8 = 24 | -16 = 240 - 16 = 226. think of few examples, solve them using pen...cross check ur answers...when accuracy becomes gud, try doing it w/o pen....got lotsa other such methods...lemme know if its helpful...if so...wud post more... hope the post helps...practice guys...practice!!!

regards maxximus===========================

Posts: 320 Join Date: Dec 2006 Location: I:Boston Heart:India Groans: 51 Groaned at 20 Times in 13 Posts Thanks: 657 Thanked 1,980 Times in 209 Posts Send a message via Yahoo to maxximus Report Bad Post Re: Concepts...total fundas!! - 31-05-2007, 02:35 AM - Add Post To Favorites Today's concept : Finding HCF n LCM of typical values. #1 : to find the HCF, LCM quickly. i know most of us wud know this....if v have few nos., 20,40,50,80,180...to find their LCMs, HCF...there's a slightly quick method... express them in prime nos. 20 = 2^2 40 = 2^3 50 = 5^2 80 = 2^4 180 =3^2

x x x x x

5 5 2 5 2^2 x 5

now to HCF, see highest power of all prime nos. that are common to all nos. 2 - 2 3- 0 5 - 1 hence hcf is 2^2 x 5 = 20 to find lcm...see highest power of all prime nos across all nos. 2 - 4 3 - 2 5 - 1 hence, lcm = 2^4 x 3^2 x 5 = 1620. #2 To find HCF and LCM of the form2222....30 times. 3333....70 times. to solve such questions... for HCF..

take hcf of no. thats being repeated...i.e. hcf of 2 & 3. i.e. 1 take hcf of no. of time these nos. are being repeated...i.e. hcf of 30 n 70...thats 10. so the hcf is 111...written 10 times. For LCM... take lcm of no. thats being repeated...i.e. lcm of 2 & 3. i.e. 6 take lcm of no. of time these nos. are being repeated...i.e. lcm of 30 n 70...thats 210. so the hcf is 666...written 210 times. #3...to find hcf and lcm of following form... 2^300 - 1, 8^250 - 1. the idea is..a^n - b^n is always divisible by a-b. so v need to find highest a-b that will divide a^n - b^n and smallest term that'll be divisible by a^n - b^n. express them in a common base. 2^300 - 1 and 2^750 -1. to find hcf... take hcf of powers i.e. hcf of 300 and 750...i.e. 150 so the hcf is 2^150 - 1. to find lcm.... take lcm of powers i.e. lcm of 300 and 750...i.e. 1500 so the hcf is 2^1500 - 1.

Questions : find hcf and lcm of: 1.2222...250 times and 8888...300 times 2. 333....120 times and 1111...400 times 3. 111...700 times and 9999...200 times. 4. HCF of 33333...200 times. and 777777.....300 times 5. 32^250 -1 & 16 ^ 100 - 1. 6. 81^100 -1 & 243 ^ 200 - 1.

7. 343^150-1 & 2401^100 - 1. 8. 125^200 - 1 & 625^120 - 1. 9. 169^320 - 1 & 32^160 - 1. for the following questns... mark 1. - stmt 1 is sufficient. 2- stmt 2 is suff. 3-both are reqd to solve the questn. 4-either is suff. 5-both insufficient. 10. what is the hcf of 5 nos., a,b,c,d,e? stmt 1 - a=72,b=4,c=6 stmt 2 - d= 8, e = 27. easy set...hope many wud get all correct... regards maxximus ++++++++++===============

Originally Posted by Rockeeze View Post originally posted by iyervani 30.05.07 One of the smaller sides of a right angled triangle is (2^2)*(3^3)*(4^4)*(5^5)*(6^6)*(7^7) . It is known that other two sides are integers.How many triangles of this type are possible. The approach mentioned was: the given number is 2^16 * 3^9 * 5^5 *7^7 a^2 = c^2 - b^2 = c-b * c+b both c-b and c+b be either odd or even ..but here both cannt be odd.. so both even to write as a product of 2 numbers we can use all powers of 2 excepth the 32th power.. so no of triangles = (31*19*11*15 -1 )/2 I did not understand the highlighted part... could someone plz explain this or any other approach to solve the ques.???? well i thought dis ques was worth a discussion in this thread vanis query was also not ans der.. i too hv the same query i think the ans should be 17*10*6*8-1/2 number of ways the samller side cn be expressed as a product of two no hi rock...thanx for bringing this to the thread... this question has to do with a concept i've already discussed on this thread...concept of factors...yeah...this question has lotsa twists n turns

attached. the final answer given by whoever answered it is correct but the xplanation has more than one mistake...seems the question was solved after seeing the correct answer...another mal-practice that must be avoided...lets have an elaborate discussion... the given number is 2^16 * 3^9 * 5^5 *7^7 a^2 = c^2 - b^2 = c-b * c+b for convenience, take c-b = t, c + b = u. both t and u shud be either odd or even ..but here both cannt be odd (since a pythagorian triplet with hypotenuse = even n rest 2 sides = odd does not exist) so both even now a^2 = 2^32 * 3^18 * 5^10 *7^14. now, since both t & u are even, 2x2 already exists in t * u. so the powers of 2 which can be floated across 2 terms to be multiplies reduces by 2. i.e. 32 -2 = 30. now, the no. of ways in which pairs can be formed are 31 * 19 * 11 * 15. but lets not forget out of these cases, there lies a case when t = u....or lets say b = 0. that particular case shud be discarded... since the factors wud be used in pairs, answr shud be half the no. of factors...so the final answer is... (31*19*11*15 -1 )/2 please feel free to revert.. with due regards maxximus ========================== Answers for yesterday's questions... 1.2222...250 times and 8888...300 times----22..50 times, 88..1500times.2. 333....120 times and 1111...400 times---11..40 times, 33..1200 times. 3. 111...700 times and 9999...200 times.----11...100 times, 99..1400 times. 4. HCF of 33333...200 times. and 777777.....300 times --- 11..100 times, lcm is diff to find... (33333...200 times. * 777777.....300 times) / 11...100 times. 5. 32^250 -1 & 16 ^ 100 - 1----2^50 - 1, 2^10,000 - 1 (nobody got this right) 6. 81^100 -1 & 243 ^ 200 - 1.---- 3^200-1, 3^2000 - 1 7. 343^150-1 & 2401^100 - 1.----7^50 - 1, 7^1800 - 1 8. 125^200 - 1 & 625^120 - 1.---- 5^120-1, 5^2400 - 1

9. 169^320 - 1 & 32^160 - 1.---- 1, prod of 2 nos. for the following questns... mark 1. - stmt 1 is sufficient. 2- stmt 2 is suff. 3-both are reqd to solve the questn. 4-either is suff. 5-both insufficient. 10. what is the hcf of 5 nos., a,b,c,d,e? stmt 1 - a=72,b=4,c=6 stmt 2 - d= 8, e = 27. answer - option 2. stmt is suffct. This was the only tricky question in the set...look at statement 2...8,27 are coprime nos. so their hcf is 1. hence, hcf of the entire set is 1. so, stmt 2 is sufficient to answer.

regards maxximus Complete Guidance for GMAT and Beyond...Click Here!! Last edited by maxximus; 01-06-2007 at 01:10 AM. Digg this Post!Add Post to del.icio.usStumble this Post! � Quote � Quick Reply Groan! Thank! Multi-Quote: Multi-Quote This Message The Following 6 Users Say Thank You to maxximus For This Useful Post: HarshaRocks (02-10-2007), hiran.prashanth (04-02-2009), kaizen_2007 (10-08-2007), mohit_ranka (02-06-2007), mr.s.k.abhi (31-10-2007), Rockeeze (01-06-2007) maxximus View Public Profile Send a private message to maxximus Visit maxximus's homepage! Find all posts by maxximus Add maxximus to Your Contacts (#130) maxximus maxximus is offline In India until 21st July Hardcore PaGaL Posts: 320 Join Date: Dec 2006 Location: I:Boston Heart:India Groans: 51 Groaned at 20 Times in 13 Posts Thanks: 657 Thanked 1,980 Times in 209 Posts Send a message via Yahoo to maxximus Report Bad Post Re: Concepts...total fundas!! - 01-06-2007, 01:11 AM - Add Post To Favorites Quote:

Originally Posted by Rockeeze View Post hi was going thru the concept of lcm n hcf type #2 222.........30 times 333..............70 times the method is fine, but cud nt find a logic for taking the lcm of 30 n 70 22......... 30 times=2*111..........30 times n 333......70 times=3*111.........70 times cud u plz explain the logic?? 111...30 times can be written as 111...10 times x 10^20 + 111....10 times x 10 ^10 + 111...10 times. hence, v can say 111...30 times is divisible by 11...10 times. similarly.. 111...70 times can be written as 111...10 times x 10^60 + 111...10 times x 10^50 +......111...10 times. hence this is also divi. by 111...10 times. so, the hcf is 111...10 times. please note that 10 times is the max no. of times 11..can be written so that it divides both 30 times n 70 times... while finding lcm, v need 11... as many times that 11...70 times as well as 11..30 times...so it shud be 111....210 times. do revert.. regards maxximus ================================ Today's concept : To solve last digit problems. Type # 1: questions of the form... 2003 x 2004 x 4235161006 x 432657178001 x 42315098002 x 423087004 concept: Last n digits of any product depends on the product of last n digits. so just multiply last n digits of each term...find the product, take last n digits of the product n multiply it with the next term...continue this for all terms. suppose v had to find last digit of the product above... so multiply last digits.

3 x 4 = 12. dont worry abt 1 in 12. just remember 2 and multiply it with next no. 2 x 6 = 12. so 2, 2 x 1= 2, 2 x 2 = 4, 4 x 4 = 16. so, the ans is 6. for last 2 digits: 03 x 04 x 06 x 01 x 02 x 04 = 76 for last 3 digits: 003 x 004 x 006 x 001 x 002 x 004 = 576. trust me, last 4 digits wont be asked...as then it becomes bulky...questions in cat are tricky... Type # 2: of the form...last digits of 432^43567. i.e base ^ power i know most of us know this concept...wud take it concisely... look for the variation in last digit of higher powers of last digit of the base...i.e. 2 here. 2^1 2^2 2^3 2^4 2^5

= >2 => 4 => 8 => 6 => 2

so we can say that 2,4,8,6 will keep repeating...no. of different digits that the last digits of higher powers can take is known as cyclicity. every digit has a cyclicity. to find last digit, find last digit of : (last digit of base) ^ (power % cyclicity of last digit) when, power % cyclicity of last digit = 0, take, (last digit of base) ^ (cyclicity of last digit) in the above example... 432^43567. 2^ (43567 % 4) = 2^3 = 8 answer. with little practice....u can do all such questions mentally. Type # 3 Find last 2 digits of 34291^201.

when the last digit is 1,9,0,5...try finding a pattern in last 2 digits...u'll get one...then solve the question accordingly... last last last last . . . . . . . last last

2 2 2 2

digits digits digits digits

of of of of

91^1 91^2 91^3 91^4

= = = =

91 81 71 61

2 digits of 91^10 = 01 2 digits of 91^11 = 91

see, we again got 91 as last 2 digits...so v can say that the cyclicity of 91 for last 2 digits is 11 -1 = 10 so the answer shud be 91.

lets take one more example... find last 2 digits of (49)^(37)^(38 )^(39)...(3700) see... last last last last

2 2 2 2

digits digits digits digits

of of of of

49^1 49^2 49^3 49^4

=> => => =>

49 01 49 01.

can v say that for all odd powers, answer wud be 01? yes! since the power of 49 is odd...the answer shud be 01. what if the question was (49)^(37)^(38 )^(39)...(3700)%100 = ? answer wud be 01%100 = 1

(49)^(37)^(38 )^(39)...(3700) % 20 = ? answer wus be 01 % 20 = 1. bcoz, to c remainder with 20, v need last 2 digits only.

(49)^(37 )^(38 )^(39)...(3700) % 10 => 49 % 10 = 9

this method might become tedious when the last 2 digits are unfriendly...like 37, 82 etc. but i have neva seen such figures appearing in cat...to solve such figures, v need euler's or binomial...ill be taking it when v discuss remainders...today's questions are based on the above concept only.. Questions for today.. 1. last digit of 3677^400 - 689^84 2. last digit of 11^11 + 12^12....1000^1000. 3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451...do i need to say....its tricky...not lengthy!! 4. no. of zeroes at the end in 1^1x2^2x3^3x...250^250. 5. no. of zeroes at the end in 1! x 2! x 3! x...25! 6. Last non-zero digit of 25! 7. Last non-zero digit of 1! x 2! x 3!...15! Find last 2 digits of... 8. 81^(371)^(372)^...(400) 9. 11 ^ (25)^(31)^(41)...(1001) 10. 7 ^ 2501. 11. 3^2537837. Hope the post helps... with warm regards maxximus ====================== hie....answers...!!

1. last digit of 3677^400 - 689^84---- 0 2. last digit of 11^11 + 12^12....1000^1000.----nobody got this right 3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451...do i need to say....its tricky...not lengthy!!----nobody got this right. 4. no. of zeroes at the end in 1^1x2^2x3^3x...250^250.----try again 5. no. of zeroes at the end in 1! x 2! x 3! x...25!----56

5-9...5 nos. one 0 each = 5 10-14...5 nos. two 0 each=10 15-19...5 nos. three 0 each=15 20-24...5 nos. four 0 each=20 25...1 no. with 6 0s = 6 total = 56 6. Last non-zero digit of 25!----try again 7. Last non-zero digit of 1! x 2! x 3!...15!----try again Find last 2 digits of... 8. 81^(371)^(372)^...(400)---81 9. 11 ^ (25)^(31)^(41)...(1001)----51 10. 7 ^ 2501.----07 11. 3^2537837.----63 kudos to ramkrishnas, jhajee...who got many correct answers. questions needed to be discussed further.... 1. last digit of 11^11 + 12^12....1000^1000 2. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451 3. no. of zeroes at the end in 1^1x2^2x3^3x...250^250. 4. Last non-zero digit of 25! 5. Last non-zero digit of 1! x 2! x 3!...15! try these questions...discuss them on the thread...few of them are a bit lengthy but very useful from practice / concept point of view...i've deliberately made them long...to let everybody understand wat happens to the last digit when a no. is multiplied with 15,20,25 etc. take ur time to solve...as i said...this set is v useful and wud bring in lotsa new concept as u solve it.. with warm regards maxximus[/quote] ========================= Answers... 1. last digit of 11^11 + 12^12....1000^1000 Ans 3 from 1^1 till 10^10, we'll get 1,6,3,6,5,6,7,4,9,0. the order will change but the set will be of these 10 digits only...just an observation!! so, we're adding the bove sum (1000-10 ) /10 = 99 times...so the last digit shud

be last digit of 47 x 99 = 3. 2. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451 Ans - 50 there's one 25 n one 2...that product will make the last 2 digits 50. now 50 x any odd no. will always leave 50 as last 2 digits...n all the other nos. are odd 3. no. of zeroes at the end in 1^1x2^2x3^3x...250^250. Ans 8125 5s... 5 ( 1+ 2+....50) = 6375 25s...25(1+2+..10) = 1375 125s...125(1+2) = 375 total = 8125 4. Last non-zero digit of 25! Ans - 5 keep multiplying last digits...not a matter of more than 90 secs. twists: last non zero digit of 14! = 2..so that of 15! shud be (odd no.)2 x 15 => 8, after excluding 0... odd no. before 2 bcoz v know the no. shud be divisible by 4...an even no. before 2 wud prohibit that last non zero digit of 19! = 6..so that of 20! shud be 2 x 2 => 4, after excluding 0. last non zero digit of 24! = 6..so that of 25! shud be 25 x (odd no.)6 => 5, after excluding 0. odd no. for the same reason. 5. Last non-zero digit of 1! x 2! x 3!...15! ans 6 same approach as above kudos to sumi,rock,saurabh...who got many correct answers!!!! questions for today.... find remainder when divided by 100... 1. 1^1 + 11^11 + 111^111 +.... thousand ones^thousand ones 2. 2^20,000 3. 1^1 + 21 ^21 + 31^31 + ....1001^1001 4. 5^5 + 15^15 + 25 ^25 +...125^125 5. 9^9 + 99^99 +.....hundred 9s ^ hundred 9s

happy solving!!!! regards maxximus ======================== hi everybody...sorry but no sorry for a deliberate late reply...am overwhelmed by the active participation out here....so decided to stay away till the heat settles....am very happy that so many posts discussed today's questions...n v had a lot of drama before managing all the correct answers... this has been a special experience...because it's for the first time that all the questions in the set have been rightly answered...i hope the trend continues... nothing wud please me more than to copy-paste solution by others using my methods suggested here...today my job looks easy...n am a happy man wud request everybody to pour in their self-discovered / short cut methods....

find remainder when divided by 100... 1. 1^1 + 11^11 + 111^111 +.... thousand ones^thousand ones ----90 The remainder when divided by 100, depends upon the last two digits... So the last two digits of the expression are 01 + 11 + 11 +....... = 01 + 999*11(last two digits of it ) = 01+89 = 90 2. 2^20,000----76 2^20000=(2^4)^5000=16^5000 Now the cyclinity of 16 for the last two digits is 5.... So the answer is 16^5= 76

3. 1^1 + 21 ^21 + 31^31 + ....1001^1001----90 (thanx for noticing missing 11^11...ya i know i have a history of typo...still...as rock said...its always safe to solve wats printed!!) last two digits of the expression is given by (01 + 21+ ..... +91 )+ (01+11 + 21 +...... +91 .... 10 times(till 991) + 01 (for 1001) 449 + 460 *9 + 01 = 449 + 4140 + 01 = 4590 so the answer is 90

4. 5^5 + 15^15 + 25 ^25 +...125^125----25 every expression gives 25 as last two digits 25 + 25 +...... 13 times....= 25*13=25

5. 9^9 + 99^99 +.....hundred 9s ^ hundred 9s----90 89 + 99*99 = 89 + 01 = 90 kudos to MOHIT, ROCK, SAURABH, SUMI, JHA for their correct answers as well as for the healthy discussions....keep up the spirit guys!!!! regards maxximus ===================== Today's concept : Mastering Data suffieciency. With no. of options for every question increased to 5, the expected answer set for D.S. is... 1. Statement 1 alone is sufficient 2. statement 2 alone is sufficient. 3. both statements together can answer the question but none of them is alone sufficient. 4. either statement is sufficient. 5. both statements are insufficient. wud like to share my approach towards D.S. first n the mother of all other ...see if statements 1 is sufficient. if yes... answer shud be 1 or 4. ............now check if statement 2 is sufficient. ..................if yes, option 4 is correct. ...........................if no, option 1 is correct. dont see if both together can answer the que. if no....answer can be 2,3 or 5 ...........check if stmt 2 is sufficient. .................if yes, option 2 is correct. dont see if both together can answer the que. ..................if no, answer shud be 3 or 5. .......................check if both 1 & 2 together lead to a unique solution. .............................if yes, option 3 is correct ....................................if no, option 5 is correct. do not forget... 1). The answer shud be unique...even 2 cases suggest that the statement is not suffiecient. 2). The answer shud be consistent...if the answer is no...it shud be always

no...if yes...always yes. 3). Both statement can give different answers...they're independent till used together...so even if the answers r different...the answer wud optn 4 if each statement satisfies above 2 conditions independently. 4). at times u'll see that question can be answered using any one option as well as both options...in such a case, answer wud be 1 or 2 and not 3. 5) the most important rule....everytime u're able to sort a D.S. question easily...check twice...there must be a catch sumwhere... Questions for today... 1. Statement 1 alone is sufficient 2. statement 2 alone is sufficient. 3. both statements together can answer the question but none of them is alone sufficient. 4. either statement is sufficient. 5. both statements are insufficient. 1. is X positive? stmt 1. x^2 - 5x + 10 = 7 stmt 2. sqrt (x) is real. 2. is X even? stmt 1. 7x + 5y = even stmt 2. 3x + 38y = odd. 3. three packages combinedly weight 60 kg. what is the weight of heaviest package ? stmt 1. one package weighs 10 kg stmt 2. one package weighs 32kg 4. is X > Y? stmt 1. 2/5 X > 3/8 Y stmt 2. 3/7X > 15/29 Y 5. If p^q = 289, (p-q)^(q/2) = ? stmt 1. p > q stmt 2. p is not = 1. 6. for above questn. stmt 1. p>q stmt 2. p is prime. 7. are sum goats not cows? stmt 1. all cows are lions stmt 2. all lions are goats.

8. is the no. of flowers in the garden = 3576876767576654544334538? st 1. no. of flowers in a row = no. of flowers in a column. st. 2 no. of rows = no. of columns. 9. for above question st 1. no. of flowers in a row = no. of flowers in a column. st 2. no. of flowers are even. 10. x = ? st. 1 x^2 - 5x + 6 = 0 st. 2 x^3 - 8 = 0 Happy solving!!! regards maxximus ========================== is X positive? stmt 1. x^2 - 5x + 10 = 7 stmt 2. sqrt (x) is real. ans. 1...bcoz both roots for st 1 are positive...so the answer wud be 'yes' always....while for stmt 2, answer wud be yes for all real vaues of x except 0. since, 0 is very much real but non-positive...so 2 is insufficient. 2. is X even? stmt 1. 7x + 5y = even stmt 2. 3x + 38y = odd. ans.2...bcoz 38y is even...so 3x must be odd for the sum to be odd. while in stmt 1, x n y are either both even or both odd...so inconsistency.. 3. three packages combinedly weight 60 kg. what is the weight of heaviest package ? stmt 1. one package weighs 10 kg stmt 2. one package weighs 32kg ans.2...if one package is weighing more than half the total...it's gotto be heaviest. 4. is X > Y? stmt 1. 2/5 X > 3/8 Y stmt 2. 3/7X > 15/29 Y to be discussed further...with xplanations... 5. If p^q = 289, (p-q)^(q/2) = ? stmt 1. p > q stmt 2. p is not = 1. to be further discussed...few of u got the correct answer but with a fallacious approach...xplanations required!!

6. for above questn. stmt 1. p>q stmt 2. p is prime. to be further discussed...few of u got the correct answer but with a fallacious approach...xplanations required!! 7. are sum goats not cows? stmt 1. all cows are lions stmt 2. all lions are goats. ans.5...due to one possible case when all lions are cows n all goats are lions..in this case the answer wud be NO. otherwise its yes...inconsistency...hence option 5. 8. is the no. of flowers in the garden = 3576876767576654544334538? st 1. no. of flowers in a row = no. of flowers in a column. st. 2 no. of rows = no. of columns. 9. for above question st 1. no. of flowers in a row = no. of flowers in a column. st 2. no. of flowers are even. nobody got these right....wud like u guys to think more... 10. x = ? st. 1 x^2 - 5x + 6 = 0 st. 2 x^3 - 8 = 0 ans.3...bcoz st1 will give 2 values....2,3 so alone not suff. st1 will give 3 values...2, other complex...so not sufficient...but using both, 2 is common...so using both options, v can answer. kudos to mohit,rock,sumi,dranzer,cat2007,apjonline,junoonmb a for their many correct answers and healthy discussions.... questions to be further discussed 1. is the no. of flowers in the garden = 3576876767576654544334538? st 1. no. of flowers in a row = no. of flowers in a column. st. 2 no. of rows = no. of columns. 2. for above question st 1. no. of flowers in a row = no. of flowers in a column. st 2. no. of flowers are even. 3. If p^q = 289, (p-q)^(q/2) = ? stmt 1. p > q stmt 2. q is not = 1. (edited...made q is not= 1, hint for all...whether its p not =1 or q not = 1, answer remains unchanged) 4. for above questn.

stmt 1. p>q stmt 2. p is prime. 5. is X > Y? stmt 1. 2/5 X > 3/8 Y stmt 2. 3/7X > 15/29 Y i hope we get them all soon...there's sum catch in every question....so come up with ur answers as well as xplanations... regards maxximus ======================== Hie...we've seen a lot happening over the set of 10 D.S questions....later reduced to 5...there have been right approaches, wrong answers as well as wrong approaches and right answers (lets hope v carry this luck to cat..!!!) few correct approaches have been given up...and few wrong approaches have been appreciated....high time i shud give a solution to the more trickier 5... i request everybody here to look at the solution very carefully... 1. is the no. of flowers in the garden = 3576876767576654544334538? st 1. no. of flowers in a row = no. of flowers in a column. st. 2 no. of rows = no. of columns. to ensure that the no. of flowers shud be a perfect square, stmt 1 is enough...try making a grid without equal rown n columns but equal no. of flowers...u'll fail... so, if the given figure is a perfect square, this stmt becomes insufficient as there are other perfect squares also. but if its not a perfect square, then this statement becomes sufficient n v get a consistent answer with this stmt that the given no. is NOT the no. of flowers in the garden. since, the given no. ends in 8, its not a perf square and thus, not the no. of total flowers in the garden..so v get a consistent answer... come to stmt 2...no. of columns = no. of rows does not necessarily mean that no. of flowers are also equal in each column n row...think...their can be jus plants with no flowers etc. try imagining...u'll understand...so this stmt can give yes as well as no as the answer...so this stmt is insufficient. so, the answer is 1. 2. for above question st 1. no. of flowers in a row = no. of flowers in a column. st 2. no. of flowers are even. stmt 1 is sufficient...as stated above. stmt 2 is insufficient as the given no. is even but there are many even nos. and the given no. of flowers may/may not be the correct value...

so, the answer is 1. i hope i was justified in re-stating this question many of us cud still not get the correct answer. 3. If p^q = 289, (p-q)^(q/2) = ? stmt 1. p > q stmt 2. q is not = 1. (edited...made q is not= 1, hint for all...whether its p not =1 or q not = 1, answer remains unchanged) 4. for above questn. stmt 1. p>q stmt 2. p is prime. well....mohit did a great job out of these questions....the answer is 5 for both the questions... important: always see if it's been mentioned whether the values are I,N,W,R or nothing has been mentioned...had there been an additional information that p,q are N, both the questions wud have been answerable... 5. is X > Y? stmt 1. 2/5 X > 3/8 Y stmt 2. 3/7X > 15/29 Y well..well...here v come...mother of all problems...see carefully. stmt 1 reduces to X/Y > 15/16. now if X = 15.5, Y = 16, answer is NO. but if X = 17, Y =16, answer becomes yes...so, this stmt is insufficient... now the important 1... stmt 2 reduces to X/Y > 35/29. since the ratio is always > 1, this statement is sufficient. what about negative values of X n Y? please understand...the form given to us is 3/7X > 15/29 Y and not X/Y > 35/29...so if v r assuming some -ve values of X n Y, they shud hold true for the main form as well. if v assume that X = -36, Y = -29, X/Y > 35/29: will hold true as the value becomes 36/29 which is > 35/29 but this gives X < Y. but are v allowed to conclude here that this statement is insufficient? NO!!...v must check for the main for given... 3/7X > 15/29 Y @ X = -36, Y = -29, becomes... -108/7 > - 15 does this hold gud??? NO!!! that means X will always be > Y if the main condition has to hold true!! so, the answer is 2 important: in inequalities, think twice before taking ratios, cancelling similar terms, sending denomainator to right side or bring numerator of right side as denominator in left side....when v do that, signs come into picture...as now 2

negative values will make a positive value...n we'll be comparing 2 positive values....whereas had v not done the shifting, both values wud have been negative n we wud have been comparing 2 negative values....n remember, when v change signs, inequality shud reverse...n when v take ratio...the inequality is not being reversed. e.g. -4 < -3 this is correct!! but, 4/3 < 1 is not correct!! (specially for u mohit bhai...) thats y...maths is beautiful...n the numbers...wonderful!!! wud like to hear from u guys with due regards maxximus ps sorry for so many bolds...but so much in the post was so much important that cudnt avoid... ========================= MORE D.S QUESTIONS: 1. is x > y? st.1 3x = 2k st.2 k = y^2. 2. is X/12 > y/40? st1. 10X > 3Y st2. 12X< 4Y 3. Are two triangles congruent? st.1 both are right trangles st 2 both have same perimeter 4. is quadrilateral ABCD a rectangle? st1. angle A = angle C = 90 st2 AB is parallel to CD 5. mohit drinks glycodin when he's upset. is he upset? st.1 he's drinking glycodin st.2 he's not drinking glycodin. 6. if x = k, is x^3 + ax^2 + bx =0? st.1 a = 0 st.2 -b = k^2 7. a dozen eggs cost 90/- in jan 1980. did a dozen egg cost more than 90/- in jan 1981? st.1 in jan 1980, the average worker had to work 5 mins to pay for a dozen.

st.2 in jan 1981, the avrg worker had to work 4 mins to pay for a dozen 8. i started for my office, half an hour late than everyday. by how many minutes was i late for the office today? st.1 i travelled at 4/5th my usual speed. st.2 i tuk a root that was 4/5th the usual distance. 9. Rock bought X flowers for Y dollars (X and Y are integers). had he purchased 20 more flowers, he wud have got all for T dollars and saved 10 cents per dozen. how many dollars did he spend earlier? (remember, questions on flowers have a history...the history continues...have a tuf time solving this one!!!) st1. T = 2 st2. he got 10% discount in second case. 10. sumi lived for less than 100 years. in which year was she born? st. 1 in 20th century, she said he was x years old in the yr x^2. st. 2 she lied while making statement 1. happy solving regards maxximus ========================= hey...seek solutions so that v move to the next concept soon...has everybody noticed these questions? MORE D.S QUESTIONS: 1. is x > y? st.1 3x = 2k st.2 k = y^2. 2. is X/12 > y/40? st1. 10X > 3Y st2. 12X< 4Y 3. Are two triangles congruent? st.1 both are right trangles st 2 both have same perimeter 4. is quadrilateral ABCD a rectangle? st1. angle A = angle C = 90 st2 AB is parallel to CD 5. mohit drinks glycodin when he's upset. is he upset? st.1 he's drinking glycodin st.2 he's not drinking glycodin.

6. if x = k, is x^3 + ax^2 + bx =0? st.1 a = 0 st.2 -b = k^2 7. a dozen eggs cost 90/- in jan 1980. did a dozen egg cost more than 90/- in jan 1981? st.1 in jan 1980, the average worker had to work 5 mins to pay for a dozen. st.2 in jan 1981, the avrg worker had to work 4 mins to pay for a dozen 8. i started for my office, half an hour late than everyday. by how many minutes was i late for the office today? st.1 i travelled at 4/5th my usual speed. st.2 i tuk a root that was 4/5th the usual distance. 9. Rock bought X flowers for Y dollars (X and Y are integers). had he purchased 20 more flowers, he wud have got all for T dollars and saved 10 cents per dozen. how many dollars did he spend earlier? (remember, questions on flowers have a history...the history continues...have a tuf time solving this one!!!) st1. T = 2 st2. he got 10% discount in second case. 10. sumi lived for less than 100 years. in which year was she born? st. 1 in 20th century, she said he was x years old in the yr x^2. st. 2 she lied while making statement 1. and sunil natraj has given an additional question...sum of 100 terms of n(n+2) do start com'n up with solutions... @sunilnatraj...ur expertise wud be helpful bro... @focused2007...u're welcome to the thread!! happy solving regards maxximus +======================== Hie...here's the solution to the last set of D.S questions...sorry for the delay!! 1. is x > y? st.1 3x = 2k st.2 k = y^2. Ans - 5...easily, neither is alone sufficient and using both, v get x = 2y^2/3 which may or may not be greater than y. e.g. y=1 => x = 2/3 hence answer wud be NO. and y = 3=> x=6, now the answer wud be YES. so v cant answer this ques. 2. is X/12 > y/40? st1. 10X > 3Y st2. 12X< 4Y Ans � 1...the question trims down to if x/y > 3/10. using st 1, we consistent YES,using st2 v dont get consistent yes or NO as st2 is X/Y<1/3..which may or may

not be >3/10. so, statement 1 is alone sufficient, st2 is not. Dont worry abt negative values....as xplained in the last set! 3. Are two triangles congruent? st.1 both are right trangles st 2 both have same perimeter Ans - 5...two such sets do exist...e.g. 3,4,5 and 2 , 4.8 , 5.2 i.e same perimeter,rt triangles but incongruent. 4. is quadrilateral ABCD a rectangle? st1. angle A = angle C = 90 st2 AB is parallel to CD Ans - 1..mohit's solution Drawing two diagonally opposite 90 degrees lines on paper, will give us a figure like the following ................C ..............7 (seven means a right angled joint, it should be taken as inverted L) ...L A Now, extending these two points A and C to create the the quardilateral ABCD, gives us a rectangle. statement 2 is NOT sufficient alone,as AB || CD, but there may not be rectangle, if BC is not || AD. So the answer is 1 5. mohit drinks glycodin when he's upset. is he upset? st.1 he's drinking glycodin st.2 he's not drinking glycodin. Ans - 2...welll...mohit knows exactly when he drinks glycodin...so he's the best person to answer this...his solution: statement 1 is NOT sufficient alone as mohit MAY drink glycodin when he is NOT upset too. statement 2 is sufficient as he is not drinking glycodin means he is NOT upset :smile:. So the answer is 2 6. if x = k, is x^3 + ax^2 + bx =0? st.1 a = 0 st.2 -b = k^2 Ans � 3 Mohit's soln..statement 1 is NOT sufficient alone, as it reduce the term to k^3 + bx, which might be and might not be equal to zero. statement 2 is also not sufficient as it reduces the term to ak^2, which again might be and might not be equal to zero. considering both statements together, we get 0 in L.H.S, hence the answer is 3 7. a dozen eggs cost 90/- in jan 1980. did a dozen egg cost more than 90/- in jan 1981? st.1 in jan 1980, the average worker had to work 5 mins to pay for a dozen. st.2 in jan 1981, the avrg worker had to work 4 mins to pay for a dozen Ans - 5...v have no idea wat happened to wages of workers. @mohit...bhai...neva asume...what if i assume that the wages definetly changed? the idea is..wages may or may not have changed...hence, 5. 8. i started for my office, half an hour late than everyday. by how many minutes

was i late for the office today? st.1 i travelled at 4/5th my usual speed. st.2 i tuk a root that was 4/5th the usual distance. Ans � 5...this is a tricky one...nobody got this right...its more abt language of the question...see...everywhere the term is 'in reference to usual stuff'...30 mins late from usual time...4/5th usual speed...4/5th usual distance...but look at the question...by how many minutes was i late for OFFICE...the reference has been changed from usual timings to office...we donno how punctual i am at my office...may be i reach 30 mins early everyday...or may be i reach 1 hr late everyday...hope its clear!!! 9. Rock bought X flowers for Y dollars (X and Y are integers). had he purchased 20 more flowers, he wud have got all for T dollars and saved 10 cents per dozen. how many dollars did he spend earlier? st1. T = 2 st2. he got 10% discount in second case. Ans � 1....nobody got this either....a very simple problem though...rock spent 2 dollars while buying 20 more flowers...so he must have spent <2 dollars earlier....right? st1 says he spent 2 dollars later...so earlier he must have spent 1 dollar only...as its given that he spent integral no. of dollars...no calculation reqd...st1 alone suffct, st2 is not..hence 1. 10. sumi lived for less than 100 years. in which year was she born? st. 1 in 20th century, she said he was x years old in the yr x^2. st. 2 she lied while making statement 1. Ans-1. Mohit gave a beautiful soln...here's wat he wrote.. the only perfect square is 1936 (of 44) in 20th century. hence sumi was born in 1936-44 = 1892. statement 2, just says that the year is not 1936, and hence its NOT sufficient alone to tell the age of sumi. so the answer is 1 kudos to mohit,rock,sumi,ankur,apjonline,jha,junoonmba who got many correct answers!! up next..remainders regards maxximus =======================================================

Remainder questions can be broadly divided into 2 types... 1). LCM based questions. e.g. a no. leaves remainders 3,2 when successively divided by 5,6...what will be the remainder when this no. is divided by 30? 2). Power based questions. e.g. remainder when (31)^[(373)^(432)] is divided by 7. we have already discussed type 1 in two parts...those who have missed it or wish to revisit the concept may use the link below. first half http://www.pagalguy.com/forum/quanti...-fundas-2.html (Concepts...total fundas!!) second half

http://www.pagalguy.com/forum/quanti...-fundas-5.html (Concepts...total fundas!!) so what we're left to discuss is type 2 mentioned above...i think there are 3 ways of solving these questions...v gotto use our own sensibility to c which approach suits where... a) Using binomial theorem b) using cyclicity with remainders c) using euler's theorem. for better understanding, i'll divide this post into 3 parts...n will discuss one approach in each part... Finding remainders using binomial theorem. (x+y)^n can be expressed as : nC0 x^n + nC1 x^(n-1)y^1 + nC2 x^(n-2)y^2 + .......... nCr x^(n-r)y^r + ..... nCn y^n concept: first term, i.e nC0 x^n is the only term that is independent of y and last term, i.e nCn y^n is the only term that is independent of x. rest, all the terms are divisible by both x & y. we'll leverage this property to solve complex problems. e.g. 28^37 % 9 = ? see...v can express it as (27 + 1 )^37 % 9. now, since 27 is a multiple of 9, the only term that'll be independent of 9 will be the last term i.e. nCn 27^0 * 1^n = 1. hence, the remainder is 1%9 = 1. important observation: we know 1^ (any damn thing...even infinity) = 1 and (-1)^(anything) = 1 or -1 with even and odd values of power respectively. so we'll try getting a form of (nD +/1)^N. so that v r ultimately left with (+/-1)^N. 342^423 % 7 = ? step1: 342%7 = 6. so the question becomes... 6^423 % 7. express 6 as 7 - 1. so the expression becomes... (7-1)^423. the only terms independent of 7 is ... -1^423 = -1.

hence, the remainder is -1 + 7 = 6.

523^325 % 7 =? => 5^325 % 7. now, i seek a remainder of 1 or 7-1=6 with a power of 5. 5^1 % 7 = 5....wont work 5^2 % 7 = 4....wont work. 5^3 % 7 = 6...will work. so, (5^3)^ (325/3) = (5^3)^ (108 ) x 5 = (-1)^108 x 5 = 1x5 =5...the remainder.

529^700000 % 7 = 4^700000 4^3%7 = 1. hence, we'll express the given expression in terms of 4^3 4^3 ^ (700000/3)...c how to save time...700000%3 = (7 + 0 + 0 + 0 + 0 + 0) % 3 = 1. hence (700000-1)/3 will be an integer...dont evaluate its value... bcoz 1^ anything = 1. hence v have 1^I x 4^1 = 1 x 4 = 4 therefore, the remainder is 4.

(31)^[(373)^(432)] % 7 = ? 31 % 7 = 3 hence, =(3)^[(373)^(432)] 3^3 = 27 = 28 -1. hence, expression shud be in terms of 3^3 = 3^3 ^ [(373^432)/3] now treat [(373^432)/3] as a new, different question... (373^432)%3 = 1. hence, the term becomes 3^3^(I) x 3^1 I = (373^432 - 1)/3 and is of the form (odd-odd) / odd = even/odd = even for

sure!! hence, the expression becomes.... (3^3)^(even I) x 3 % 7 = (-1)^even integer x 3 = 1 x 3 = 3 hence, remainder = 3.

case when divisor is a large no. in such questions, v try to reduce power by increasing the value of base and bringing it close to a multiple of divisor. e.g. 2^35 % 61 = ? v know 2^6 = 64 hence, (2^6)^5 x 2^5 2^6 % 61 = 3. hence, 3^5 x 2^5 = 3^4 x 3 x 2^5 =3^4 % 61 = 20. 3x 2^5 = 96, 96%61 = 35. hence, 20 x 35 % 61 = 700%61 = 29, the reqd answer. hope the concept helps...will shortly post the remaining 2 parts as well... regards maxximus Complete Guidance for GMAT and Beyond...Click Here!! Last edited by maxximus; 09-06-2007 at 12:46 PM. Digg this Post!Add Post to del.icio.usStumble this Post! � Quote � Quick Reply Groan! Thank! Multi-Quote: Multi-Quote This Message The Following 26 Users Say Thank You to maxximus For This Useful Post: anshul14 (26-06-2007), atshubha (31-07-2007), daniket (26-06-2009), gautamgomzi (08-10-2007), gmat_delhi2007 (07-09-2007), harneet9 (21-06-2008), HarshaRocks (0210-2007), iyervani (09-06-2007), ketce (22-06-2007), kondapalli (06-09-2007), kooldud (16-09-2007), mba.yodha (10-09-2008), mohit_ranka (10-06-2007), mukg (1406-2009), neha_tyro (25-10-2008), Rockeeze (09-06-2007), Rohan Koshy (31-07-2009), saurabhgulati (10-06-2007), scorpion_girl (20-06-2007), sgx100 (13-06-2007), SUMI_08 (10-06-2007), TargettCAT (07-09-2008), varshita (09-06-2007), vivek.anand (13-06-2007), vyomb (09-06-2007), willrock (30-06-2008) maxximus

View Public Profile Send a private message to maxximus Visit maxximus's homepage! Find all posts by maxximus Add maxximus to Your Contacts (#323) ============================= Finding remainders using cyclicicty with remainders: This approach is useful when the divisor is small or at times when it is a factor of 100. 3^327%7 = ? 3^1 3^2 3^3 3^4 3^5 3^6 3^7

% % % % % % %

7 7 7 7 7 7 7

= = = = = = =

3 2 6 4 4x3 % 7 = 5 5 x 3 % 7 = 1 1 x 3 % 7 = 3

remainder with first power is same as remainder with 7th power...hence v can say that cyclicity in remainders is 7-1 = 6. so, 327 % 6 = 3, hence, effectively, the remainder is 3^3 % 7 = 6 326^524 % 9 =? 326 % 9 = 2 hence, 2^524 now, 2^1 2^2 2^3 2^4 2^5 2^6 2^7

% % % % % % %

9 9 9 9 9 9 9

= = = = = = =

2 4 8 7 5 1. 2

hence cyclicity in remainder = 7-1 = 6. 524 % 6 = 2 hence, the remainder is 2^2 % 9 = 4.

81^502 % 100 81^1 % 100 = 81 81^2 % 100 = 61

81^3 81^4 81^5 81^6

% % % %

100 100 100 100

= = = =

41 21 01 81

hence, cyclicity = 6-1 = 5. 502 % 100 = 2. so the reqd remainder is same as that with 81^2 = 61. similarly...this method can be effectively used when remainders are other factors of 100. viz, when the factors are 20,25,50,100...knowing last 2 digits wud suffice knowing the remainders...we've already discussed this concept while discussing cyclicity... important: dont try this method when the divisors are complex...viz 37,73 etc...the cylicty wud come very late n calculations will grow cumbersome...when divisors are complex, there must be sum other catch in the question...look for that catch...e.g. last type discussed in the binomial method... regards maxximus ============================= Finding remainders using Euler's theorem. (special thanks to junoonmba for this) This method is very useful when the divisor and dividend are relatively prime numbers... step 1: To calculate euler's no. of a divisor. euler's no. can be practically taken as cyclicity in remainders by a divisor.. to find euler's no, express the divisor in terms of prime factors... 100 = 2^2 x 5^2. powers of the prime nos. have no significance...its jus the prime no. that matters... euler's no (e for convenience) = divisor x (1-1/first prime factor) x (1-1/second prime factor) x ... (1-1/last prime factor) so, for 100, e = 100 x (1-1/2) x (1-1/5) = 100 x 1/2 x 4/5 = 40. that means e for 100 = 40. or, in other words, 100 divisor will definetly show a cylicity of 40 in the remainders. whenever the power of a relatively prime no. will be a multiple of 40, the expression wud show a remainder 1 with 100. e.g. 3^120 % 100 = ?

we know e for 100 = 40. 3 n 100 are relatively prime nos. hence, 3^40 % 100 = 1. hence 3^120 % 100 = (3^40)^3 % 100 = 1^3 = 1.

7^100 % 45 = ? 45 = 3x3x5 e for 45 = 45 x (1-1/3) x (1-1/5) = 24 hence, 7^24 % 45 = 1 hence, 7^100 % 45 = 1^4 x 7^4 % 45 = 2401 % 45 = 16, the required answer... with this, we have finished the conceptual part of remainders...now what we seek is practice...we need to solve a wide variety of problems to improve our reflexes while choosing the best method for solving a question.... a sincere request to everybody around...please drop in the most difficult questions on remainders u've eva encountered...let's solve them thru various approaches and discuss our solutions...for the coming few days...lets know remainders inside out... regards maxximus +++=======================

Hi puys, Its been many days since I last posted on PG, but I was occupied with unavoidable commitments , and now it seems, I have lot of catching up to do... Firstly solution to QQAD # 49 (Sorry for the delayed post) As its mentioned in the question that the age of the twins is a teen non-prime no. hence it can be either 14, 15, 16 or 18.... Probable Ages Sum Product Ages Raveena Thought 14, 14, X 28 + X 196X 7,7, 4X (not satisfy add. criteria) 16, 16, X 32 + X 256X 2,2, 64X (___ DO____)

18, 18, X 36 + X 324X 2,2, 81X (___ DO____) ...............................................3,3 , 36X (___ DO____) 15, 15, X 30 + X 225X 3,3, 25X (satisfies all criteria for X=1) Thats why the ages are 15, 15, 1.... I think i am clear enough :smile: Regards, Mohit PS - There have been some doubts regarding this question. Kindly read the conversation between me and vineetvijay, from here onwards..... it will help, grasping the problem.. http://www.pagalguy.com/forum/quanti...s-day-165.html (CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions) P P S : @ Maxximus bro, Please invite problems from other puys, only regarding to the current-concept-in-consideration..... If there will be many posts comprising different concepts, we will not be able to learn the concept thorouly, which will defy the Aim of this thread. "Democracy, is an anarchy, having multiple dictators" --Unknown

Moi Blog: http://mohitranka.wordpress.com Moi Trainee PaGal now... Digg this Post!Add Post to del.icio.usStumble this Post! � Quote � Quick Reply Groan! Thank! Multi-Quote: Multi-Quote This Message The Following 2 Users Say Thank You to mohit_ranka For This Useful Post: Rockeeze (10-06-2007), rosh! (14-06-2007) mohit_ranka View Public Profile Send a private message to mohit_ranka Visit mohit_ranka's homepage! Find all posts by mohit_ranka Add mohit_ranka to Your Contacts (#345) mohit_ranka mohit_ranka is offline has no status. Trainee PaGaL mohit_ranka's Avatar Posts: 92 Join Date: Aug 2006 Location: Bhilwara Age: 23 Groans: 12

Groaned at 5 Times in 4 Posts Thanks: 121 Thanked 59 Times in 40 Posts Send a message via Yahoo to mohit_ranka Report Bad Post Re: Concepts...total fundas!! - 10-06-2007, 06:40 AM - Add Post To Favorites Quote: Originally Posted by junoonmba View Post Solve this...really nice There are 1000 soliders with a no. on their T Shirts (1,2,3.....999,1000) standing on a circular track. Man with no.1 is carrying a sword in his hand. He kills man at no. 2 and pass on sword to no. 3. This goes on until we have only one solider on the track. What will be the no. on his T-Shirt? If n is the last number on the T-shirt (or n is the total number of people), then we observe the following pattern n ___________________ last surviour T-shirt number 1 ___________________ 1 2/3 _________________ 1/3 4/5/6/7 ______________1/3/5/7 2^n/..... /2^(n+1)-1 ____1/3......./2^(n+1)-1 so for n=1023, the surviour will be 1023. and from back tracking from 1022 to 1000, we get for n=1000, the surviour will be 1023-2*23 = 1023-46 = 977 Hope thats correct!! Regards, Mohit Tip: For these kind of questions, where you can find the answer by mechanical calculations, but n is very large, try to find a relation in the answer and n, or in other words try to generalize the problem for any value of n and substitute the original large n in the generalized relation to get the answer..... key thing is to realize that in which question, generalization will work... "Democracy, is an anarchy, having multiple dictators" --Unknown

Moi Blog: http://mohitranka.wordpress.com Moi Trainee PaGal now... Last edited by mohit_ranka; 10-06-2007 at 10:34 AM. Reason: Added Tip Digg this Post!Add Post to del.icio.usStumble this Post! � Quote � Quick Reply Groan! Thank! Multi-Quote: Multi-Quote This Message The Following 7 Users Say Thank You to mohit_ranka For This Useful Post: junoonmba (10-06-2007), martinet (23-06-2008), maxximus (10-06-2007), neha_tyro (25-10-2008), priyom (27-07-2007), Rockeeze (10-06-2007), vivekr (10-06-2007) mohit_ranka View Public Profile Send a private message to mohit_ranka

Visit mohit_ranka's homepage! Find all posts by mohit_ranka Add mohit_ranka to Your Contacts (#346) junoonmba junoonmba is offline out on the town and pigging out!!!!! Addicted PaGaL SJMSOM Mumbai Posts: 1,060 Join Date: May 2006 Location: delhi Age: 26 Groans: 56 Groaned at 56 Times in 31 Posts Thanks: 719 Thanked 782 Times in 351 Posts Report Bad Post Re: Concepts...total fundas!! - 10-06-2007, 10:36 AM - Add Post To Favorites @Mohit_Ranka Yes, you are correct and nice explanation ........................... Some food for thought... 1.Remainder of 26^16 divided by 125 2. Remainder of 3^94 divided by 125 3.Reamainder of 2^13 divided by 25 4. Remainder of 10^14 divided by 37 5 2^1990%1990 Some phew.. questions for practise..... ============================================ Ok...euler's cannot be applied here since 7 and 28 are NOT relatively prme... so... 7%28 == 7 7^2 %28 == 21 7^3%28 == 7 7^4 %28 == 21 and so on...this pattern repeats... so 7 to the power of an even number gives the remainder as 21.. so my ans is 21 ======================= Quote: Originally Posted by vyomb View Post puys help me in solving this stsmt a and b are to be analyzed and answered 1 if qns can be answered by using one of the stmt alone but nt by using the other stmt 2if the qns cna be answered by using either of the stsmt alone 3if the qns can be answered by using both the stmts together 4if the qns cannot be ansered on the basis of any of the stmts Zakib spends 30% of his income on his children's education,20% on recreation and 10% on health care.The corresponding percentages for Supriyo are 40%,25% and

13%.Who spends more on children's education? A.zakib spends more on recreation than supriyo B.supriyo spends more on helthcare than zakib hi vyomb... the answer wud be 1. i.e. using statement B alone...see the solution... for expenditure on child..., .3Z > .4S or, Z/S > 1.3333 for the expense of A to be more... now use statements... A says that .2Z > .25S => Z/S > 1.25 which may or may not be > 1.3333...hence st A is insufficient... st B says .13S >.1Z ....hence, Z/S < 1.3 therefore, this is definetly less than 1.3333.. so stB will consistently give one answer...NO. hence, answer is 1, using statement B only... feel free to revert regards maxximus ============================ hie...here's a less than 3 mins solution to 2^1990 % 1990 please see carefully...u can pick a new concept from thsi post... take divisor as 10 x 199 last digit of 2^1990 = 4. hence remainder when 2^1990 is divided by 10 is 4. now lets take 2^1990 % 199 e for 199 = 198 that mean 2^1980 % 199 = 1. hence, 1 x 2^10 % 199 is effective remainder with 199. i.e 1024%199 = 29, easily so with 199, remainder is 29 with 10, remainder is 4 so, v r lookin for a no. that leaves 29 rem with 199 and 4 with 10. guys who have been following this thread wud know how to solve this part quickly no. wud be (29 + 199k) such that (29 + 199k) % 10 = 4

(29 + 199k) %10 can be reduced to... 9 + 9k = 4,14,24... smallest integral value is k=5. hence the no. v r looking for is 29 + 199 x 5 = 1024 1024 % 1990 = 1024 hence, the answer is 1024.

now that v have solved all the questions with short, practical approaches...can v practice few more? solve these... 2^2!^3!^4!...100! % 1. 7 2. 9

3. 5^32 % 5000 4. 2^60 % 130 5. 7^50 % 1001 regards maxximus ============================= Maxximus bhai.........Thoda gyaan humari taaraf se :-) From Number system:Concept 1:(N^p-1)-1%p gives remainder----0 where p is prime number nd n,p are relativily coprime................. Concept 2:(p-1)!+1 %p gives remainder-------0 where p is prime number Questions for practise--------1 99^6-1%7 2 456^18%19

3 (741^198 -192)%199 4.100!%101 5 (28!+233)%899 ======================== hey sumi! here goes the concept of inverse euler number:: say you have to calculate the remainder of the question:: 4^15/17 you can very easily go the traditional way and calculate but the concept of inverse euler will prove handy and will save whole lot of time. 4^16 / 17 will be 1 (why????) 4^ 15= 4^16*4^-1 thus the question will become 4^16*4^-1/ 17 which is same as 4^-1/17(why?????) FUNDA:: say if you have N^-b/p then you look for a number "A" such that N*A/p will be 1. After that the above equation can be written as A^b/p. See in the above case:: what is the smallest number which should be multiplied with 4 so that remainder will be 1. hmmmmmm..pondering over it and you will see 13 is the number. thus 4*13/17 is 1. hence equation will reduce to 13^1/17 which is nothing but 13. cross check by going it the novice way. some more for practice:: 8^828/167 31^78/123 5^26/29 regards!! marijuana_user. Want deep insight on GMAT!! Visit:: www.expertsgloba ===========================

well the formula gives e.n of 100 as 40,but what i have learnt and what i have seen is that it is better to break up the bigger number in terms of its prime and then calculate the euler number.I have explained the same case for 77,which by formula gives e.n as 60 but if u break it up in terms of 6 and 11 then you get the answer as 30.Cross check it by calculating the remainder for 3^32/77 or say 12^36/77 hi marijuana...its great to c u active on the thread... abt the discrepancy in the euler's....wud like to throw some light...while pondering over euler's...even i tried coming up with short-cuts...e.g. i know with 3, 100 wud show cyclicity in remainder after 20 only n not 40..still i believe both methods are fine...n v can stick to basic euler's to avoid confusion... 1). Technically speaking....even if v come up with some re-arrangement...v shud not (a euphemism for v font have the authority to) say that euler's no. has changed...that wud cause confusion as technically euler's no. is wat v get from x(1-1/p1)(1-1/p2)... where p1 and p2 are prime nos. so when v take 25 n 4...v aint following eulers as these aint prime nos. hence, it's be better to say that euler's no. can be taken as 20 for ease of calculation. 2). now when u say bigger nos, there can be many set of bigger nos, for ex. 100 can be broken as 50,2 20,5 etc....each wud give a different e...though i'd take it for granted that when u said take bigger nos, they shud be mutually prime...infact highest powers of each prime no. taken independetly.. 3). its safe to use euler's relation...remember, when v take e as 40, instead of 20 for 100....v r only taking a multiple of 20...v are not taking a factor of 20...so even if 20 gives correct answer, there's no doubt 40 will always give correct answer. great work marijuana....nothing wrong with ur method...its only going to ease the calculations...n the answers shud always be same using conventional eulers or the method suggested by u... regards maxximus =========================== Hey max! well nice to c u 2 man! a fauji in d past! great ! m a fighter pilot in the past ! merit out ! luck had its way. anyways i just wanted to share this:: i have been following PG for a long time now and all this time that i followed it i was in college.So this is an advice for college goers:: sometimes one tends to get carried away while being online because of so many things happening on PG(it is this feature of PG which makes it out of the world and so special) and this tends to waste whole lot of time.So i request you all to keep a schedule.Dont give preference to Pg over your self study.Once you are through your self study,browse PG and share the garguntuan amount of knowledge.Follow certain threads and dont try to glance here,there and everywhere.like FOLLOWING THIS THREAD SHOULD BE PART OF YOUR SCHEDULE TILL THE VRY END(excluding the time for self study).Also some times valuable info like directors interview and all is available on PG so that becomes a must read. For working people PG is the panacea to all probs(bowing down).As it comes to the rescue,coz one can browse through it endlessly. Hope it will help!

I promise to stick to this thread till the very end. well what i observed:: 1. Most of the people active on this thread are from DELHI(sumi,Junnon,Myself,Max). 2. So why dont plan a meet on this sunday??? ===========================