Number Theory Solution

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MULTIPLE CHOICE QUESTIONS INTRODUCTION TO NUMBER THEORY Total marks 40. Each right answer will be awarded 4 and wrong answer will attract a penalty of -1 marks. Time 15 min. ……………………………………………………………………………………..

1. Small trees are to be planted in rows such that each row contains equal but odd Number of tree. If there are total of 3500 plants then the no of possible way is 1. 24 2. 8 3. 16 4. 112 Solution: 3500 = 2^2 * 5^3 * 7^1 As each row must contain odd no of tree so ans is equal to total of odd divisor = (3+1)*(1+1) = 8. hence 2 is ans. 2. A number of statements are given below with respect to two numbers 2835 and 2800. Chose the correct staement among the following: a) 2835 and 2800 have equal number of divisor b) 2835 has a total of 18 distinct factor c) 2835 and 2800 have equal number of odd divisor d) 2800 has a total of 23 even factors e) 2835*2800 have 41 distinct facors 1. b and d 2. a ,c and e 3. b, c and e 4. a and b Solution: ans is 1. As all other statement are incorrect 2835 = 3^4 * 5 * 7 and 2800 = 2^4 *5^2 *7 Hence 2835 has 18 distinct factors (all odd) And 2800 has 28 distinct factors, 5 odd and 23 even factors. 3. 32400 is decomposed into N pair of factor such that each pair when Multiplied together yields 32400 itself. Then the maximum value of N is: 1. 37 2. 38 3. 39 4. 41 Solution: 32400 = 2^4 * 3^4 * 5^2. Hence possible no of such pair is Given by 0.5*[(4+1)*(4+1)*(2+1) +1] = 38. Hence 2. 4. The no 111111111111 is perfectly divisible by: 1. 11 2. 37 3. 407 4. all of the above Solution: If no of one is even => divisible by 11 If no of one is multiple of three=> divisible by 37. Hence also divisible by 37*11=407 .Ans is 4.

5. What is the remainder when 234*235*236*237*238*239 is divided by 120? 1. 119 2. 1 3. 0 4. 2 Solution: any six consecutive no is divisible by 6! So it is also divisible By 5! or120. Hence the ans is 3. 6. Let a three digit number be in form of xyz such that 2x is arithmetic mean of y and z , also y is arithmetic mean of x and z ,then the no can be 1. 131 2. 357 3. 471 4. 253 Solution: Ans is 2 as it fulfills above condition. 7. If the number 78 in decimal base system is represented as 243 in some other base Notation then 442 in the other base notation is eqivalent to: 1. 122 in decimal system 2. 172 in octal system 3. 7A in hexadecimal system 4. all of the above 5. none of these Solution: Ans is 4. 78 in decimal system is 243 in base 5 system. 442 in base 5 system is 122 in decimal, 172 in octal and 7A in hexadecimal. 8. How many zeros are ther at the end of 126! 1. 25 2. 30 3. 31 4. 29 Solution: Zero is obtained by multiplication of 2 and 5. Total no. of 2 in 126! Is greater than total no of 5 in 126!. Hence No. of zeros is equal to no. of times 5 comes in 126! The value is equal to (126/5) + (126/25) + (126/125) = 25 + 5 +1 = 31. Ans is 3. 9. Assume n to be a positive integer and p and q are prime number. It is also given That n is greater than 2 and p, q is greater than 6 then which of the following statement is true: a) Seventh power of n substracted by n ( n^7-n ) is necessarily divisible by 42. b) Square of p substracted by square of q ( p^2 - q^2 ) is necessarily divisible by 48 c) (p-1)!+1 is divisible by p

d) (pq-1)!+1 is divisible by pq 1. a , b and d 2. a and c 3. c and d 4. a, b and c Solution: n^7-n is divisible by 6 and 7 so is also divisible by 42 Given n is greater than 2. p^2-q^2 is necessarily divisible by 24 but not by 48. c is true but d is false. Hence 2 is the answer. 10. There are 16 ways in which students of a class can be grouped such that each Group contains equal number of student. Then the minimum no of student in the Class is: 1. 120 2. 180 3. 150 4. 80 Solution: Answer is 2. As total no of distinct factor of 180 is 16 excluding 180 and 1.