what is the remainder when we divide 125! by 10^31
4
1
0
5

• 125! have 31 zero
  10/10=0 is the remender
• 9 years agoHelpfull: Yes(9) No(3)
• ANS:0
  highest power of 10 in 125! is 31 i,e (125/5)+(125/25)+(125/125)=31
  125!=(10^31)(somevalue) / 10^31 remainder is 0
• 9 years agoHelpfull: Yes(8) No(1)
• 125!=1*2*....10*....125
  10^31=10*10....
  10/10=reminder 0
  like this 1*2*...10*..125/10*10*...
  the 10/10 reminder 0
  so the reminder of the whole term will be 0
  
• 9 years agoHelpfull: Yes(5) No(4)
• factorial of 125 is 0,
  so when 0 /anything is zero,
  hence zero is the remainder
• 9 years agoHelpfull: Yes(3) No(15)
• total num of zero at the end of 125!=125/5+125/25+125/125
  =31
  when 125!/10^31 rem will be zero
• 9 years agoHelpfull: Yes(3) No(0)
• the shortcut method to find the no, of zeroes in N! is dividing N by 5 and its higher powers till we get 1 as quotient. just get the quotient and don't consider the remainder if not perfectly divisible
  similarly
  125 / 5 = 25
  125 / 5^2 = 5
  125 / 5^3 = 1
  so adding it up we get 31
  and 10^31 has 31 zeroes so perfectly divisible, hence remainder is 0
• 7 years agoHelpfull: Yes(3) No(0)
• 1 ... as example 125/1000 leaves a remainder 1
• 9 years agoHelpfull: Yes(2) No(2)
• 125! have 31 zero
  so 125/10=5
• 9 years agoHelpfull: Yes(1) No(5)
• 125!=5!.5!.5!
  we know that 5!=120
  i.e a number 120 divides by 10=0
  ans:0
• 8 years agoHelpfull: Yes(1) No(0)
• Actually the cyclicity is 3:
  
  (2^0)/7 = 0 R 1
  (2^1)/7 = 0 R 2
  (2^2)/7 = 0 R 4
  (2^3)/7 = 1 R 1
  (2^4)/7 = 2 R 2
  (2^5)/7 = 4 R 4
  ...
  
  The pattern is such that the remainder is 4 for every third one and we know there are 201 numbers between 0 and 200. Since 201 is a multiple of 3, same exp
• 9 years agoHelpfull: Yes(0) No(1)
• ans:0
  bcz 125! %10 will lead to 0.
• 5 years agoHelpfull: Yes(0) No(0)