2024 COMC B1

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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2024 COMC B1

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#1 h Nov 4, 2024, 3:51 PM

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For any positive integer number $k$ , the factorial $k!$ is defined as a product of all integers between $1$ and $k$ inclusive: $k!=k\times{(k-1)}\times\dots\times{1}$ .
Let $s(n)$ denote the sum of the first $n$ factorials, i.e.
$s(n)=\underbrace{n\times{(n-1)}\times\dots\times{1}}_{n!}+\underbrace{(n-1)\times{(n-2)}\times\dots\times{1}}_{(n-1)!}+\cdots +\underbrace{2\times{1}}_{2!}+\underbrace{1}_{1!}$ Find the remainder when $s(2024)$ is divided by $8$

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#2 h Nov 4, 2024, 3:55 PM

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Note that most of the terms of $s(2024)$ are multiples of $8$ i.e. starting with $4!$ , everything is $0\pmod 8$ . The answer is just $3!+2!+1!\equiv \boxed{1}\pmod 8.$

This post has been edited 2 times. Last edited by megahertz13, Nov 4, 2024, 3:57 PM

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#3 h Apr 2, 2025, 11:12 PM

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$3!+2!+1! \equiv 1 mod 8$

This post has been edited 2 times. Last edited by EVKV, Apr 2, 2025, 11:13 PM

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