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What is the value of $2019! \pmod{7}$?

I guess it's $0$? Because $$2019! = 2019\cdot2018\cdot2017\cdot ...\cdot7\cdot6\cdot...\cdot1$$ There's $7$ and also numbers that has $0$ remainder when divided by $7$, times the other numbers equals $0$.

Can someone correct me if I'm wrong? I'm still not sure this is the answer though. Thanks

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    $\begingroup$ All good. $2019!$ is zero mod $n$ for any $n \le 2019$ (and many more numbers besides). $\endgroup$ – TonyK Feb 14 at 13:35
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    $\begingroup$ Well you are right, by definition x=0 mod (m) iff x = k*m for some k. Since as you said 2019! = 7*k , then is 0 mod 7 $\endgroup$ – JoseSquare Feb 14 at 13:35
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    $\begingroup$ Correct, the answer is 0. This is simply because $2019!$ is a multiple of $7$. $\endgroup$ – Minus One-Twelfth Feb 14 at 13:36
  • $\begingroup$ Thanks for your help guys! $\endgroup$ – Godlixe Feb 14 at 13:45
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    $\begingroup$ The follow-up question of how many powers of 7 are in 2019! is a classic question that you may enjoy. $\endgroup$ – Jim Ferry Feb 14 at 15:46
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Yes, $$2019! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times ... \times 2019$$ $$ =7 \times (1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 8 \times 9 \times ... \times 2019), $$ so $2019!$ is a multiple of $7$ and the remainder when $2019!$ is divided by $7$ is $0,$ and therefore we write $$2019! \equiv 0 \pmod 7.$$

Note that we do not need to compute $2019!$ -- a number with thousands of digits -- to make this determination.


As pointed out in the comments, indeed all natural numbers up to and including $2019$ divide $2019!,$ by a similar argument. Also, $7$ divides $2019!$ many times (exactly how many is a different question) because of the factors $14, 21, 28, ...$ in $2019!.$

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