Find the smallest positive integer value such that for every positive integer greater than 1, is divisible by , but not by .
If there is no such positive integer value , then enter 0.
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Fact-1: One can see that for every natural number n , ( n × n ) ! = ( n 2 ) ! is divisible by n 3 .
Fact-2: x m i n = 1 doesn't work.
Whatever value you try to assign to x m i n > 1 , let it be y , when a will assume the same value and it surely will, ( x a ) ! will be then equal to ( a 2 ) ! and, by Fact-1 , be divided by ( a 3 ) .
So, such x m i n doesn't exist.
Hence, the answer is 0 .