<s>[INST] If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$ If $p$ is a prime other than 2 or 5, prove that $p$ must be one of the forms $10k + 1$, $10k + 3$, $10k + 7$, or $10k + 9$
-The section we are covering is on the division algorithm, although I am unsure as to how to utilize it in this problem. Any help is appreciated.
 [/INST] So you wish to express $10k + c$ as $bq + r$ with $0 \leq r < b$. If $r = 0$ and $10k + c \neq b$, that means $10k + c$ and $b$ share a common prime factor, and if $b$ is the prime then $10k + c$ is composite.
Try it with $b = 2$, which you know is a prime number. Then:


*

*$10k + 0 = 2(5k) + 0$

*$10k + 2 = 2(5k + 1) + 0$

*$10k + 4 = 2(5k + 2) + 0$

*$10k + 6 = 2(5k + 3) + 0$

*$10k + 8 = 2(5k + 4) + 0$


Since $r = 0$ in each of these cases and we know 2 is prime, this must mean that these $10k + c$ are composite.
Very similarly you take care of $10k + 5$ with $b = 5$.
This leaves only $10k + 1$, $10k + 3$, $10k + 7$ and $10k + 9$ as potentially prime.
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