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

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The discussion focuses on calculating the sum \( A = \sum_{k=1}^{91}(k! \times k) \) and finding \( A \mod 2002 \). A key observation is the simplification of \( k! \times k \) to \( (k+1)! - k! \). Participants note a typo in the original expression, which is acknowledged and corrected. The conversation highlights the importance of accuracy in mathematical expressions, especially when using mobile devices for posting. Overall, the thread emphasizes the process of modular arithmetic in factorial series.
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$A=\sum_{k=1}^{91}(k!\times k)$

find $ A $ MOD 2002
 
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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 $A = 2001 \pmod{2002}$.
 
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Euge said:
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 :$k!k=k!(k+1-1)=(k+1)!-k!$
 
Albert said:
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.
 
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