if $n_{3}\equiv1\pmod3$ and $n_{3}\mid7$ then $n_{3}\in \{1,7\}$

Question

It rather easy/basic question and I feel ashamed asking it but I can't figure it out.

If I know that $n_{3}\equiv1\pmod 3$ and $n_{3}\mid 7$, how to calculate $n_3$? I know that the answer is $n_{3}\in \{1,7\}$, but why?

Similarly, if we know that $n_{7}\mid 3$ and $n_{7}\equiv1\pmod 7$ why $n_7=1$?

Answer 1

Well if $n_3$ divides $7$, then $n_3$ must be either $1$ or $7$, since $7$ is prime. The first condition doesn't change that.

The same logic applies for your second part. Perhaps you could elaborate on what you're struggling with?

Answer 2

Are you saying that you do not know what "congruent to 1 mod 3" means? A number is congruent to 1 (mod 3) if and only if it is of the form 3k+ 1 for some integer k (the remainder when divided by 3 is 1). The positive numbers that are congruent to 1 (mod 3) are 3(0)+ 1= 1, 3(1)+ 1= 4, 3(2)+ 1= 7, 3(3)+ 1= 10, 3(4)+ 1= 13, etc. The negative numbers that are congruent to 1 (mod 3) are 3(-1)+ 1= -2, 3(-2)+ 1= -5. 3(-3)+ 1= -8, etc. None of those negative numbers divides 7.

The only numbers that divide 7 are 1 and 7. 1 is congruent to 1 mod any number and 7= 3(2)+ 1 is congruent to 1 mod 3, so the solution set is {1, 7}.

The only numbers that divide 3 are 1 and 3. 3 is not congruent to 1 (mod 7) but 1 is, so the solution set is {1}.