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Algebra Level 5

Find the number of solutions for natural numbers n ≤ 999 n \le 999 such that

( 4 − 2 1 ) ( 4 − 2 2 ) ( 4 − 2 3 ) . . . ( 4 − 2 n ) (4- \frac{2}{1})(4- \frac{2}{2})(4- \frac{2}{3})...(4- \frac{2}{n})

is an integer.


The answer is 999.

Hoo Zhi Yee by Hoo Zhi Yee , 26, Malaysia

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1 solution

Murugesh M
Nov 14, 2014

The product upto n terms can be written as (2n)!/(n!)^2 = 2n C n , which is an integer for all positive integral values of n.

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