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I'm faced with a difficult problem that I can't prove.

The problem is stated as:

Find the remainder when $(1! \times 1) + (2! \times 2) + (3! \times 3) + \cdots + (100! \times 100)$ is divided by $101$.

I found a pattern which is $(1! \times 1) + (2! \times 2) + (3! \times 3) + \cdots + (n! \times n)$ is exactly $n$ when in modulus $n + 1$, I couldn't proof it so I need some help, thanks!

Sayan Dutta
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Dean Tee
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    $\sum_{r=1}^n r\cdot r! = \sum_{r=1}^n (r+1)!-r! = (n+1)!-1 \equiv -1 \pmod {n+1}$. – D S Mar 06 '24 at 07:41
  • @DS It does answer my question, thank you so much! You should have make an answer instead of a comment so I can flag your answer as correct – Dean Tee Mar 06 '24 at 07:58
  • No, this is a duplicate, and duplicates should not be answered. Instead, they should be closed with a link to the original question. – D S Mar 06 '24 at 08:00
  • I am not sure how closing as duplicate works, but I think you would have seen a yes/no beside my comment? – D S Mar 06 '24 at 08:01
  • It really is, thanks! – Dean Tee Mar 06 '24 at 08:04

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Since $101$ is a prime no and as it does not appear even once until $100! . 100$ . So we will divide the sum of the series by $101$ to directly find the remainder. The sum can easily calculate by writing $S_n =\sum n .n!$ Which can be split and written as sigma of $(n+1 -1)×n!$ Which on opening becomes $n+1! - n!$.so taking the sum of the obtained term from $n = 1 \to n=100 $we get the sum as $101!-1$ which on being divided by $101$ gives $-1$ as remainder. So $-1 +101= 100$ is the remainder.

Sisanta Chhatoi
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