sum of the digits in the product (16^100)*(125^135)

• (16^100)*(125^135) = ((2^4)^100)*((5^3)^135) = (2^400) * (5^405)
  (2^400)*{(5^400)*(5^5)} = (10^400)*(3125) = 312500000......(400 times)
  so the sum of the digits are 3+1+2+5 = 11
• 9 years agoHelpfull: Yes(36) No(1)
• (16^100)*(125^135)= (2^400)*(5^405)
  which means (2*5)^400*(5^5)
  3+1+2+5+0+0+0..........=11
  
  Ans= 11
• 9 years agoHelpfull: Yes(23) No(2)
• 2^400)*(5^405)=3125*10000000..........
  there for sum of digits is 11
• 9 years agoHelpfull: Yes(3) No(0)
• 6^n=6(as last digit)
  5^n=5(as last digit)
  therefore, 6+5=11
• 9 years agoHelpfull: Yes(3) No(8)
• mariyam...can you please explain your answer in brief...because i m not able to understand after the second step???
• 9 years agoHelpfull: Yes(0) No(0)
• cycle of 6 always end with 6 & cyle of 5 always ends with 5
  so sum of digit =11
• 9 years agoHelpfull: Yes(0) No(0)
• 16^100=2^400
  125^135=5^405
  2^400*5^400*5^5=10^400*3125=312500000000000000000000000000000000000000000000000000000......
  sum of the digits=3+1+2+5=11
  
• 9 years agoHelpfull: Yes(0) No(0)