Number of trailing zeros of n! in base b(b=$$p^m$$, where p is a prime number) is for $$k\ge1$$ $$\frac{1}{m}\left(\Sigma\left[\frac{n}{p^k}\right]\ \right)$$
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Number of trailing zeros of n! in base b(b=$$p^m$$, where p is a prime number) is for $$k\ge1$$ $$\frac{1}{m}\left(\Sigma\left[\frac{n}{p^k}\right]\ \right)$$
Educational materials for CAT preparation
