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.