Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7

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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)

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4 Answers

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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 $9\pmod{10}$

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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.

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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 $7$.

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No product of two elements in $\{1,5,9\}\subset{\mathbb Z}_{10}$ is in $\{3,7\}$.

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