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

Thread starter stamenkovoca02
Start date Mar 17, 2021
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Divisibility

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The discussion centers on determining the remainder of (1!+2!+3!+...+100!)^2 when divided by 5. It is noted that 5 divides evenly into factorials starting from 5! and higher, while lower factorials like 4! and 3! do not contribute to divisibility by 5. The impact of squaring the sum on the overall divisibility is questioned. Participants are encouraged to explore the problem further rather than seeking a direct answer. The conversation emphasizes the importance of understanding factorial properties in relation to divisibility.

Mar 17, 2021

stamenkovoca02

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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Mar 17, 2021

cbarker1

Gold Member

MHB

You have to show some type of attempt. We can't provide the answer, but we can guide you to the answer.

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