Maximum value of n for which 224! is divisible by 224^n

• we know that prime factor of 224 = 2*2*2*2*2*7 = 2^5*7
  so we have to find how many 2's and 7's in 224!
  
  no of 2's in 224! is ....
  224/2+224/4+224/8+224/16+224/32+224/64+224/128=
  112+56+28+14+7+3+1=221
  
  so no of 2^5 in 224! is
  221/5=73 (because here power of 2 is 5)
  
  now no of 7 in 224! is
  224/7+224/49=32+4=36
  
  for these type of problem we always choose which have least value ,
  
  so final ans is 36....
• 7 years agoHelpfull: Yes(18) No(1)
• 1*224=224
  2*112=224
  3*73 =224
  4*56 =224
  7*32 =224
  8*28 =224
  so answer is 8
• 7 years agoHelpfull: Yes(4) No(5)
• answer is 36
• 7 years agoHelpfull: Yes(4) No(0)
• 4 tiime
  224,112x2,56x4,23x8
• 7 years agoHelpfull: Yes(2) No(5)
• ans is 6
  224X1,112x2,4x56,8x28,7x32,16x14
• 7 years agoHelpfull: Yes(1) No(0)
• 224=2^5*7
  So we have to find no of 2^5 and 7 in 224!,we will choose whichever is less.
  So,no of 2's in 224! is 221,we want 2^5,so 221/5=((2^5)^44)*2,so we will take 44 since it is the maximum power of 2^5.Now maximum power of 7 in 224! is 36 which is less than 44. So answer is 36.
• 7 years agoHelpfull: Yes(1) No(0)
• I think 8 bcaz 2+2+4=8 and this is divisble by 8.(divisible rule)
• 7 years agoHelpfull: Yes(0) No(5)
• I think answer is 75.
  digits between 1-100 sum 17 = 2 (89,98)
  digits between 101-200 sum 17 =3 (188,179,197)
  digits between 201-300 sum 17 = 4
  and so on
  we have
  2+3+4+5+6+7+8+9+10+11=65
  
  I might be wrong. please correct me!
• 7 years agoHelpfull: Yes(0) No(3)
• What are the options ??
• 7 years agoHelpfull: Yes(0) No(0)
• answer is 36
• 7 years agoHelpfull: Yes(0) No(0)
• HI, guys I need help where I can get removed amcat questions from maths?
• 7 years agoHelpfull: Yes(0) No(0)
• 36 is answar
• 7 years agoHelpfull: Yes(0) No(0)