Latest Elitmus Aptitude Question SOLUTION: 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 .

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(#M40153801) ELITMUS Numerical Ability question Number system Keep an EYE

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 .

Asked In Elitmus Shamik Bhattacharya (8 years ago)

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praveendas  (8 years ago)

max value of x is n..

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Vikas Jha  (8 years ago)

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

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Anand Kumar Sharma  (8 years ago)

(5!)n=(5n)*(4n)*(3n)*(2n)*(1n)
is divisible by 5x.
=> n must multiple of x.
=> n=Kx, where K=0,1,2,3,….,∞.

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shilpa  (7 years ago)

x should be equal to n

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Shamik Bhattacharya  (8 years ago)

Sorry... It should be (5n!)^n instead of (5!)^n...

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