Latest CAT Aptitude Question SOLUTION: N= 202 x 20002 x 200000002 x 20000000000000002 x 200000000....2 (31 zeroes) The sum of digits in this multiplication will be:

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(#M40111603) CAT Numerical Ability question number theory Keep an EYE

N= 202 x 20002 x 200000002 x 20000000000000002 x 200000000....2 (31 zeroes) The sum of digits in this multiplication will be:

Asked In CAT cut (9 years ago)

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NESAB JAFAR  (9 years ago)

answer is 75

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kaushal kishor  (9 years ago)

answer will be 20
sum of digits in 202=4 it has 1 zeros
sum of digits in 20002=4 it has 3 zeros
sum of digits in 20000000=4 it has 7 zeros
similarly in 5th number will have 31 zeros
so finally sum of digits in this multiplication is 4+4+4+4+4+4=20

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Panna Lal Patodia  (9 years ago)

Sorry for slight mistake in the previous answer.
(1) 202 * 20002 = 4040404, Note 4 appears 4 times. Sum of digits is 16.
(2) 202 * 20002 * 200000002 = 808080808080808, eight appears 8 times, so sum is 64.
Continuing in the same way, N contains 3232.... 32 times.
As sum of 32 = 3+2 = 5. Sum of digits of N is 5*32 = 160

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Panna Lal Patodia  (9 years ago)

(1) 202 * 20002 = 4040404, Note 4 appears 4 times. Sum of digits is 16.
(2) 202 * 20002 * 200000002 = 808080808080808, eight appears 8 times, so sum is 64.
Continuing in the same way, N contains 3232.... 32 times. So, sum of digits of N = 32*32 = 1024.

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Ashima Sabharwal  (9 years ago)

202*20002=4*101*10001=4*1010101
200000002*(1010101*4)=(2*100000001)*(1010101*4)=(8*101010101010101)
So,
N=2^5*(101*10001*100000001*10000000000000001*100000000000........(31 zeroes)1)
first multiplication (101*10001), sum of digits is 4
second multiplication 8
third multiplication 16
fourth multiplication 32
so,
N=(2^5)*32
N=32*32
N=1024

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