Summing the factorials

Algebra Level 5

Let $\displaystyle a_n = n!(n+1)^2$ and $\displaystyle S=\sum_{n=0}^{94} a_n$ .

Find the remainder when $S$ is divided by 97 .

Notation : $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

The answer is 95.

1 solution

Mark Hennings
Jun 18, 2016

Since $a_n \; = \; (n+1)^2 n! \; = \; (n+1) (n+1)! \; = \; (n+2)! - (n+1)!$ we deduce that $S \; = \; \sum_{j=0}^{94} a_j \;= \; 96! - 1! \; = \; 96! - 1$ Since $97$ is prime, $96! \,\equiv\, -1$ modulo $97$ by Wilson's Theorem, and hence $S \,\equiv\, -2$ modulo $97$ . The remainder is $95$ .

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Brilliant Mathematics Staff - 4 years, 12 months ago

Sir, how remainder can be negative ?

Aditya Sky - 4 years, 12 months ago

The congruence expression $a \equiv b$ modulo $n$ simply means that $n$ divides $a-b$ . There is no restriction on the signs of $a$ and $b$ . Of course, any integer $a$ is congruent to some remainder $r$ , with $0 \le r \le n-1$ .

Mark Hennings - 4 years, 12 months ago

Got it .. Thanks Sir :)

Aditya Sky - 4 years, 12 months ago

Summing the factorials

The answer is 95.

1 solution

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