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

• answer is 75
  
• 9 years agoHelpfull: Yes(1) No(4)
• 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
• 9 years agoHelpfull: Yes(1) No(5)
• 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
  
• 9 years agoHelpfull: Yes(1) No(0)
• (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.
• 9 years agoHelpfull: Yes(0) No(0)
• 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
• 9 years agoHelpfull: Yes(0) No(0)