Divisibility of (1!+2!+3!+...+100!)^2 by 5

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The discussion focuses on determining the remainder of the expression (1! + 2! + 3! + ... + 100!)^2 when divided by 5. It is established that 5 divides evenly into factorials starting from 5! and higher, which influences the overall sum. The challenge lies in understanding how the squaring of the sum affects divisibility by 5. Participants are encouraged to explore their reasoning and provide attempts to solve the problem rather than seeking direct answers.

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1.The remainder when dividing (1!+2!+3!+...+100!)^2 by 5 is?
5 divides evenly into 5!, 6!, 7!, ..., 100!. It would also divide evenly into things like 2!*8! or 20!*83!, but not 4!*3!
but whether then ^2 affects the rest?
And what is answer?Thanks
 
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