Count them all!!

For how many positive integers less than 300 is n 5 n^5 - n 2 n^2 divisible by 25?


The answer is 71.

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1 solution

Kartik Sharma
Sep 25, 2014

n 5 n 2 = n 2 ( n 3 1 ) {n}^{5} - {n}^{2} = {n}^{2}({n}^{3} - 1)

So for 25to divide this expression either n 2 {n}^{2} or n 3 1 {n}^{3}-1 have to be divisible by 25.

n 2 {n}^{2} will be divisible by 25 for all n = 5 x for all integers x. But, n < 300, so x =< 11. So, there are 59 such integers.

Now, for n 3 1 {n}^{3}-1 to be divisible by 25, either (n-1) or n 2 + n + 1 {n}^{2}+n+1 . Therefore, there are 11 more integers which we get from (n-1). Now, n 2 + n + 1 {n}^{2}+n+1 cannot be a multiple of 25. This is because product of 2 numbers can never end in digit 4 or 9.

Therefore, there are (11 + 59)= 70 possible integers.

But the main thing here is that it has one more solution i.e. n=1. This is because 25 divides (1-1=0) also.

Therefore , the answer is 70 + 1= 71 \boxed{71}

Copied from.........JOMO-6 :P ?

Krishna Ar - 6 years, 8 months ago

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No, I didn't copy. No plagiarism!!!!

Kartik Sharma - 6 years, 8 months ago

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