How to Find the Remainder of a Modulo Operation in Factorial Series?

[Albert1]

$A=\sum_{k=1}^{91}(k!\times k)$

find $ A $ MOD 2002

 

[Euge]

Spoiler

Since $k!k = k! ((k+1)-1) = (k+1)! - k!$ for all $k$, $A$ telescopes to 92! - 1. Since $2002 = 2 \cdot 7 \cdot 11 \cdot 13$, $92!$ is divisible by 2002 and hence $A = 2001 \pmod{2002}$.

 

[Albert1]

Spoiler

Since $k!k = k! ((k+1)-1) = (k+1)! - k$ for all $k$, $A$ telescopes to 92! - 1. Since $2002 = 2 \cdot 7 \cdot 11 \cdot 13$, $92!$ is divisible by 2002 and hence $A = 2001 \pmod{2002}$.

very nice !
a typo :$k!k=k!(k+1-1)=(k+1)!-k!$

 

[Euge]

very nice !
a typo :$k!k=k!(k+1-1)=(k+1)!-k!$

Thanks. I use my phone to post answers here, and sometimes the keyboard doesn't function properly. I will make the correction.

 

[CellCoree]

I would approach this problem by first breaking down the equation into smaller parts. The first step would be to simplify the summation by expanding it using the factorial function. This would result in $A = 1 + 2(1!) + 3(2!) + 4(3!) + ... + 91(90!)$.

Next, I would use the fact that $n!$ is divisible by $2002$ for all values of $n \geq 2002$. This means that for all terms in the summation where $k \geq 2002$, the value of $k!$ can be replaced with $0$ in the mod 2002 calculation. This simplifies the equation to $A = 1 + 2(1!) + 3(2!) + 4(3!) + ... + 91(90!) + 0 + 0 + ... + 0$.

Then, I would use the concept of modular arithmetic to further simplify the equation. I would break down each term in the summation by finding its remainder when divided by 2002. This would result in $A = 1 + 2 + 6 + 24 + ... + 0 + 0 + ... + 0$.

Finally, I would use the properties of modular arithmetic to simplify the equation even further. I would group the terms into pairs and use the fact that $n(n+1) \equiv 0 \mod 2002$ for all values of $n \geq 2002$. This would result in $A = 1 + 2 + (6 + 24) + (120 + 720) + ... + 0 + 0 + ... + 0$.

In the end, I would have $A \equiv 1 + 2 + 0 + 0 + ... + 0 \equiv 3 \mod 2002$. Therefore, the remainder when $A$ is divided by 2002 is 3.