Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7 Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7.
I have that such an integer n has n=3 or 7(mod 10) but don't know where to go from there.
Then show that there are infinitely many prime numbers n with n=3 or 7 (mod 10)
 A: HINT:
First of all, any integer$(N)$ ending in $3$ or $7$ will not be divisible by $2,5$(why?) 
So, the factors of $N$ must be $\equiv1,3,7,9\pmod {10}$
Observe that $1\cdot1 \equiv1, 9\cdot9\equiv1, 1\cdot9\equiv9 \pmod{10}$
So, the product of the primes(or their powers) $\equiv1,9\pmod{10}$ will be $\equiv1$ or $9\pmod{10}$ 
A: Every prime number is congruent to $1,3,7$, or $9$ modulo $10$. Here is a a multiplication table modulo $10$
\begin{array}{c|cccc}
   \times & 1 & 3 & 7 & 9 \\
   \hline
   1 & 1 & 3 & 7 & 9 \\
   3 & 3 & 9 & 1 & 7 \\
   7 & 7 & 1 & 9 & 3 \\
   9 & 9 & 7 & 3 & 1 \\
\end{array}
Note that every integer ending in $3$ or $7$ had at least one divisor that ends in $3$ or $7$. Since a number can only have a finite number of divisors, it follows that at least one prime divisor must end in $3$ or $7$.
Since $\gcd(10,3) = \gcd(10,7) = 1$, Dirichet's theorem says that every arithmetic progression of the form $10n+3$ and $10n+7$ will contain infinitely many prime numbers.
A: Maybe "prime" is a Red Herring here.  Show more generally: if a number ending in $3$ or $7$ is factored in any way (prime or not) then at least one of the factors ends in $3$ or $7$.
A: No product of two elements in $\{1,5,9\}\subset{\mathbb Z}_{10}$ is in $\{3,7\}$.
