for what maximum value of x can (7^n)! be completely divided by 7^x

• no. of 7's in (7^n)! = 7^n/7 + 7^n/7^2 + 7^n/7^3 + ... + 7^n/7^n
  
  = 7^(n-1)+ 7^(n-2)+ ... + 1
  = [7^(n-1)- 1]/ 6
  
  for (7^n)! to be completely divided by 7^x
  x(max) should be equal to [7^(n-1)- 1]/ 6
• 10 years agoHelpfull: Yes(15) No(3)
• 7^(n-1)+7^(n-2)+7^(n-3)+...7^0
  summation= ((7^n)-1)/6
• 10 years agoHelpfull: Yes(4) No(1)
• (7^(n-1))+(7^(n-2))+(7^(n-3))+......upto 7^0
• 10 years agoHelpfull: Yes(2) No(1)
• guys please tell me value of x please don't give equations , iss ko to koi bhi nikal le ga....
  
• 10 years agoHelpfull: Yes(1) No(2)
• the expansion for (7^n)! would look like:
  (7^n)! = (7^n).((7^n)-1).((7^n)-2)....((7^n)-r).....3.2.1 ,by defn of factorial
  To be divided completely by 7, there are two types of terms in the series:
  1. that exhibiting powers of 7 e.g. 7,7^2,7^3,7^(n-1),7^n etc. and,
  2. others multiples of 7 e.g. 14,21,28,36,54 etc.
  If we try to add up no. of 7's present as a factor in above equation, then depending upon n value we get series as: 1,8,57,400,2801.... as VIVEK shown,for n=1,2,3...where each term is a summation of GP as below: Therefore, nth term is:
  term = 7^(n-1) + 7^(n-2) + 7^(n-3) +.....+ 7^(n-r) +.....+ 7^2 + 7^1 + 7^0
  which is ultimately the required value of x. Hence, using GP summation formula we get, ((7^n) - (7^0))/(7-1) = ((7^n) -1)/6
• 10 years agoHelpfull: Yes(1) No(0)
• we say no......
• 10 years agoHelpfull: Yes(1) No(1)
• for n=1
  7!/7^x
  x=8
  
  n=2
  (7^2)!/7^x
  x=8
  
  n=2
  (7^2)!/7^x
  x=8
  
  n=2
  (7^2)!/7^x
  x=8
  
  n=3
  (7^3)!/7^x
  x=57
  
  n=4
  (7^4)!/7^x
  x=400......
  
  depend upon n value if specify....
  
• 10 years agoHelpfull: Yes(0) No(4)
• avinash can u tell me among these options which one is correct
  1>
  ((7^(n-1))-1)/6
  2>
  n^3
  3>
  (n+1)^2
  
• 10 years agoHelpfull: Yes(0) No(1)
• for x=n , (7^n)! is completely divisible by 7^x.
• 10 years agoHelpfull: Yes(0) No(2)
• guys i think answer will be n^3..what u say?
• 10 years agoHelpfull: Yes(0) No(1)
• hey guys y confusing me.
  pls give the ans with explanaton
• 10 years agoHelpfull: Yes(0) No(0)