If 5 n is divisible by 5 x Find the maximum value of x in terms of n Sorry I

Ok.. I will try to solve this question for (5n!)^n instead of (5!)^n as told by shamik.
I don't know if you know this but if we have to find no of 5 in 5n! it'll be like this..
(5n/5)+(5n/5^2)+(5n/5^3).....(5n/5^n)
for eg.. no of 5 in 10! will be 10/5=2
no of 5 in 25! will be (25/5)+(25/25)=6..
so for n=1,2,3,....
no of 5 will be.. 1,2,3,4,6,7,8,9,10,12... and so on..
Now.. there is one thing to note that 20! will have 4 no of 5,, but 25! will have 6 no of 5.. Similarly 45! will have 10 no of 5 but 50! will have 12 no of 5's..
So the series will be like
(1+2+3.... )-(5+11+17...) whose sum will be
{n(n+1)/2}-5/2{10+(n-1)6}
Solve this you'll get n^2-29/2*n-10.. (Check)
Hence x={n^2-29/2*n-10)^n
9 years agoHelpfull: Yes(5) No(1)
(5!)n=(5n)*(4n)*(3n)*(2n)*(1n)
is divisible by 5x.
=> n must multiple of x.
=> n=Kx, where K=0,1,2,3,….,∞.
9 years agoHelpfull: Yes(3) No(3)
x should be equal to n
8 years agoHelpfull: Yes(2) No(0)
Sorry... It should be (5n!)^n instead of (5!)^n...
9 years agoHelpfull: Yes(1) No(0)

If (5!)^n is divisible by 5^x. Find the maximum value of x in terms of n.
Sorry , I can't recapitulate the options .