Aspire Faculty ID #16455 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Find the remainder when $X = 1! + 2! + 3! + \cdots + 100!$ is divided by $240$.

Solution

For $n \ge 6$, 
$n!$ is divisible by $240 = 2^4 \cdot 3 \cdot 5$. 
 Thus only first 5 factorials matter: 
 $1! + 2! + 3! + 4! + 5!$ 
 $= 1 + 2 + 6 + 24 + 120 = 153$

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