Marvelous Mental Arithmetic Tricks

New mental arithmetic techniques/tricks AMARNATH MURTHY CHIEF ENGINEER (ELEX & TELCOM) ONGC, AHMEDABAD, INDIA

Fifth root of a 10 digit numbers which are perfect fifth powers. First we have to list and memorise the fifth power of numbers from 1 to 9. 1 , 32, 243, 1024,

3125, 7776, 16807, 32768, 59049

The beauty is the unit digit is the same. A^5 Ξ A mod 10. ( A =1 to 9) Consider an up to 10 digit number as a fifth power of a two digit number. If the number has n digits n <= 10, consider the number formed by the n-5 MSDs ( most significant digits). Let it be N Find numbers r and r +1 such that r^5 < N < (r +1)^5. Let the unit digit of the given number be ‘a’. Then digit ‘r ‘ followed by digit ‘ a’ is the required fifth root. Example : find the fifth root of 656356768 . The unit digit of the fifth root is 8. Leaving first five digits we have 6563. 5^5 = 3125 < 6563 < 7776 = 6^5 hence the tens digit is 5 The answer is 58. In fact one need not know the digit 2nd to 5th . i.e. if the perfect fifth power number given is 6563 _ _ _ _8 , still one can find the fifth root.

Fifth root of a 15 digit number Finding the fifth root of a 15 digit number involves the knowledge of some more concepts in mathematics. i.e. modular arithmetic.

Number N 1 2 3 4 5 6

N^5 1 32 243 1024 3125 7776

N^5 mod 9 1 5 9 7 2 9

N^5 mod 11 1 10 1 1 1

10

7 8 9 Multiple of 11

16807 32768 59049 -

4 8 9 -

10 10 1 11 **

Just remember that 1. Mod 9 values for 1 , 8 and 9 are the same (1,8,9) 2. Mod 9 value for 2 is 5 , 5 is 2 , 4 is 7 and 7 is 4. 3. Mod 9 values for 3, 6 and 9 is 9. This creates the problem and one can not uniquely find the number ( the fifth root). Mod 11 values for 3, 6, 9 come to the rescue with some intelligent estimation. Example1: 49420005843968 Step1. Make groups of five digit numbers from the right. 4942

00058

43968

Step 2. The first group gives unit digit 8. Step3. Consider the last group 4942.

It lies between 3125 and 7776 i.e. 5^5 and 6^5. Hence the third digit is 5. What remains is to find the middle digit. Let it be A. The required number is 5A8. Step4. Find 49420005843968 mod 9 i.e. the digital root of the number. Go on summing the digits till one arrives at a single digit number. Or by casting out 9 from 49420005843968 we get 8. 8 corresponds to 8 from the table. Hence 5A8 mod 9 = 8 which gives 5 + A + 8 = 17 => X = 4. And the answer is 548. Note1: The question that may arise in one’s mind is, as to why the sum is chosen as 17 and why not 26. The answer is simple; A is a single digit number. In case of a number like 9A9 with mod 9 value as 8 the obvious choice for the sum is 26. 9+A+9 =26 which would have given A =8. Note2: In this case the mod 9 value is other than 9 hence the need to examine mod 11 does not arise. Let us take another example. Example 2: 58602385427607 Step1: group the numbers as below. 5860

23854

27607

The first group on the right gives unit digit as 7. Step 2. 5860 lies between 3125= 5^5 and 7776= 6^5. This gives the last digit as 5. Let the middle digit be A Then the number is 5A7. Step3

Find the mod 9 value for 58602385427607. This comes out to be 9. 9 corresponds to 3, 6 or 9 in the table. There are three possibilities 5A7 mod 9 = 3
A = 0 or 9 suggesting

507 or 597

5A7 mod 9 = 6
A = 3 suggesting 537 5A7 mod 9 = 9
A = 6 suggesting 567. From here there are two methods First method ( an intelligent guess).This requires a good number sense. Consider 5860. It lies almost half way through 3125 and 7776.( slightly on a higher side at about 60%). The intuitional guess suggests that 567 should be the answer. Note; The most significant 4 digits would be closer to 3125 in 507^5 and to 7776 in 597^5. 507^5 = 33499613519307, 597^5 = 75835343042757 Second method: This method does not leave scope for any confusion or ambiguity. Consider the mod 11 value of 58602385427607 Sum (1st digit +3rd digit +... ) – Sum (2nd digit + 4th digit + ...) (7 +6 +2+5 + 3+0+8) –(0+7+4+8+2+6+5) = 31 – 32 = -10
-10 +11
1 With mod 9 value as 9 and mod 11 value as 1 the two possibilities are ( 3 and 9) ;507 and 567 which clearly suggests 567 as the right choice for 5860 is more close to 7776 than to 3125. Note: In case the number is a multiple of 11 so would be the fifth root and to solve for A the middle digit, use the equation

first digit + third digit = A+11 Finding out the seventh root of 14 and 21 digit numbers. Number N N^7 Mod 9 Mod 11 1 1 1 1 2 128 2 7 3 2187 9 9 4 16384 4 5 5 78125 5 3 6 279936 9 8 7 823543 7 6 8 2097152 8 2 9 4782969 9 4 Multiple of 11 11 First one has to memorise this table which is not very difficult after some practice. No need for learning it by heart. 1. The unit digits are 1
1, 2
8,3
7, 4
4, 5
5, 6
6, 7
3, 8
2, 9
9 The same as those for cubes. 2. The beauty of mod 9 values is one has N^7 mod 9 Ξ N, except for numbers 3,6, and 9. 3. The mod 11 values are unique. Hence considering mod 11 values suffices. Note: It is advisable to check for mod 9 values first. If it does not come out to be 9 then the solution is immediate else either intelligent guess might work or in the worst case one has to go for mod 11 values. For up to 14 digit number one need not even need or remember the 7th power of 1 to 9. Example1: 2207984167552 The unit digit is 8. Let the tens digit be A. The number required is A8.

2207984167552 mod 11 can be obtained as the difference of alternate digit ( odd – even) sums. 2207984167552 mod 11 = 9 which corresponds to 3 A8 mod 11 = 3 which gives A = 5 ( simply by testing for 18,28,38,48,58,68,...etc) And the seventh root is 58. For more than 15 digit one has to remember the 7th power of 1 to 9. Example2: 14652484962038521963 Step 1: group the numbers in set of 7 digits from he right. 146524 8496203 8521963 The fist group gives unit digit as 7 146524 lies between 78125= 5^7 and 279936 = 6^7 Hence third digit is 5 Let the middle digit be A then the required numbers is 5A7. N = 14652484962038521963 N mod 9 = 7 and the number corresponding to it is 7 Hence 5A7 mod 9 = 7. 5 +A +7 = 16 giving A = 4. And the required 7th root is 547.

The technique works for cube roots as well.

9th root of 27 digit numbers. Number N 1 2 3 4 5 6 7 8 9 Multiple of 11

N^9 1 512 19683 262144 1953125 10077696 40353607 134217728 387420489 -

N^9 mod 9 1 8 9 1 8 9 1 8 9 -

N^9 mod 11 1 6 4 3 9 2 8 7 5 0

Unit digits directly follow. The mod 9 values are not useful. Mod 11 values are all different. Hence it is easy to uniquely identify the number. For an 18 digit number it is not required to even memorise the 9th powers of numbers 1 through 9.

Number N 1 2 3 4 5 6 7 8 9 Multiple of 11

N^11 1 2048 177147 4194304 48828125 362797056 1977326743 8589934592 31381059609 -

N^11 mod 9 1 5 9 7 2 9 4 8 9 -

N^11 mod 11 1 2 3 4 5 6 7 8 9 0

For up to a 22 digit number perfect 11th power, one need not require to know or remember the 11th power of 1 to 9 to find out the 11th root. And the beauty is that the mod 11 values are identical to the number itself. Example1: 313726685568359708377 Unit digit is obviously 3. Let the tens digit be A A3 mod 11 = 313726685568359708377 mod 11 = 7 The 11th root is 73. Example 2: 30155888444737842659 Unit digit is 9. Let the tens digit be A. A9 mod 11 = 30155888444737842659 mod 11 = 4 That gives A = 5 and the 11th root is 59.

There are easy vedic methods to multiply numbers close to 100 or 1000 etc. By choosing appropriate bases. But these method loses its advantage for numbers in between like 46* 58. Here is another method which is general and more suitable for better estimates. If no rounding off is applied the method gives correct answers. We can use the identities 1. A*B = (A+x)* (B – B*x/(A+x)) 2. A*B = (A-x)* (B + B*x/(A-x)). The following example will explain it better. 489*512 ...... choose x = 11 = 500*( 512 - 512*11/500) = 500*(512 -5632 /500) ---- >> using short cut for multiplication by 11 =500*(512- 11.264) = 500*(500.736)  (divide by 2) = 250368 (halve the right member and remove the decimal point) Alternately using the second formula 512*489 choose x = 12. =500*( 489 + 489*12/500) =500*(489 + 5868/500)  using short cut for multiplication By 12 =500*( 489 +11.736) =500*500.736 (halves the right member and removes the decimal point.) =250368 This method is more suitable for rounding off estimates for multiplication of say numbers like 4.89 * 5.12 up to two decimal places. Let us try some rounding off techniques and check the error. 4.89*5.12 Step one find out 489*512

489*512 = 500 * (512- 512*11/500) = 500*(512 -5632 /500) = 500*(512 -5600 /500) roughly 500* (512- 11.2) 500*500.8 =25040 Approx. Value = 25.040 The correct value is 25.0368 Example1: 3.84* 8.57 = 400* ( 857 – 857*16/400) choosing x = 16 =400* (857 – 857/25) multiply by 4 and place the decimal accordingly = 400* (857 -34.28) = 400* 821.72 = 400* 822 328800 32.88 The right answer is 400* 82172/10000 =32.9088.

example 3: 988*456 Choose x = 12 = 1000*(456 – 456*12/1000) = 1000* ( 456 – 5472/1000) = 1000* (456 – 5.472) = 1000*450.528 = 450528

Multiply two digit numbers with 9 AB > AA THEN AB*9 =

A_ X_ (10-B) where X is obtained by the equation

A + X + ( 10-B) = 9K, K = 1 or 2. If AB < = AA then 9AB = (A-1)_ X_ (10-B) and X is obtained by (A-1) + X + ( 10-B) = 9K, K = 1 or 2. Example1: 48*9 48 > 44 hence hundreds digit is 4 10-8 = 2
unit digit is 2 Let the middle digit be A 4A2 is the number and A = 9 – 4-2 = 3 48*9 = 432. Example2: 87*9 87< 88 hence hundres digit is 8-1 = 7 Unit digit = 10-7 = 3, Let middle digit be A The number is ‘7A3’ in which 7 + A + 3 = 18
A = 8 87*9 = 783.

Multiplying by 9 (general method)

Example 1: 548*9 Technique: Unit digit is 10-unit of multiplicand. One borrowed from tens digit mandatorily. For rest of the digits borrow one from the left if needed and then subtract the left Unit place 10- 8 Tens place 7-4 = 3 Hundreds place 14-5 = 9 One borrowed from 5 hence Last digit is 4 The answer is 548*9 = 4932. Example 2: 666*9 Unit place 10 – 6 =4 Tens place 15-6 = 9 Hundreds place 15-6 = 9 Last digit 6-1 = 5 Answer is 5994 = 666*9 Example 3: 12345678*9 10-8=2, 7-7 = 0, 7-6 = 1, 6-5=1, 5-4 =1, 4-3=1, 3-2=1, 2-1=1, and 1-0 = 1 111111102