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$5^{2001} + (27)!$ is divided by 8. Could someone please help me solve this. I managed to show that $5^{2001} \equiv 13(mod8)$ but now I am stuck and don't know what to do to show the remainder when it is $27!$

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    $\begingroup$ $27! = 1\times2\times\cdots \times 8\times \cdots \times 27$ $\endgroup$ – Strants Feb 4 '15 at 4:57
  • $\begingroup$ I still need more help, I don't know how to use that to find the remainder $\endgroup$ – marcus Feb 4 '15 at 5:00
  • $\begingroup$ If $27!$ is a multiple of $8$, then what is $27!\mod 8$? $\endgroup$ – Strants Feb 4 '15 at 5:02
  • $\begingroup$ 27! is congruent to 0 mod(8). Is that correct? $\endgroup$ – marcus Feb 4 '15 at 5:12
  • $\begingroup$ Yup! So what does that tell you about $5^{2001} + 27!\mod 8$? (Remember that $13 \equiv 5 \mod 8$) $\endgroup$ – Strants Feb 4 '15 at 5:19
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$$(\text{mod}\ 8)$$

$$5^2\equiv25\equiv8\cdot 3+1\equiv1$$

Then $$5^{2001}\equiv(5^2)^{1000}\cdot 5\equiv1\cdot 5=5$$

Since $27!$ is multiple of $8$ we get $$5^{2001}+27!\equiv5+0\equiv5.$$

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  • $\begingroup$ so is the remainder 5? I am still a little confused? $\endgroup$ – marcus Feb 4 '15 at 5:39
  • $\begingroup$ @marcus Yes, $0\leq 5<8$ and $5^{2001}+27!\equiv 5(\text{mod}\ 8)$. $\endgroup$ – Pp.. Feb 4 '15 at 5:41

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