# ShareMyScreen/ProjectEuler/0012

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The count of factors of n can be found using , where *p* is the prime factors of n and *k* is the length of *p*, or in J `*/ 1 + _ q: n`. It is generally common to see some form of brute-force approach with these problems, so my initial idea was to find `i` where `A000217(i)` is the answer. I generally reach for the Do-While construct because it is fairly easy to use. `u^:v^:_` executes `u` so long as the boolean condition `v` returns a 1.

A000217 =. -: * >: NB. closed form to find +/ >: i. n cond =. 500 > [: */ 1 + _ q: A000217 >:^:cond^:_ [ 8 NB. the first 7 triangular numbers can't be it so let's start at 8 12375 A000217 12375x NB. the answer 76576500