How to prove that $x + y \equiv 4\pmod 6 $

Question

Prove that if x and y are integers such that $x \equiv 3 \pmod{12}$ and $y \equiv 7\pmod{18}$, then $x + y \equiv 4\pmod6$

I tried making the equations into algebraic equations. So,

$x\equiv3\pmod{12}$ becomes $x=3+12p$ for $p\in \mathbb{Z}$

$y\equiv7 \pmod{18}$ becomes $y = 7 + 18q$ for $q \in \mathbb{Z}$

Then, $3+12p+7+18q=4+6k$ for some $q \in \mathbb{Z}$

Then I simplified it: $6p + 3q = k-1$

This is as far as I did.

Can anyone help me in solving this problem? I am confused. Thank you so much in advance.

$6$ divides both $12$ and $18$, so $x\equiv3 \pmod{12}$ $\implies x\equiv 3\pmod{6},$ and $y \equiv 7 \pmod{18}$ $\implies y \equiv 7 \pmod{6}$, then $x+y = \dots$ — dxiv, Oct 01 '21 at 06:35

score 2 · Answer 1 · answered Oct 01 '21 at 05:51

2

You can continue where you left off:

$$x=12m+3,~~y=18n+7\\ x+y=6(2m+3n+1)+4.$$

answered Oct 01 '21 at 05:51

lone student

score 1 · Accepted Answer · answered Oct 01 '21 at 06:14

Alternative approach:

Prove that if x and y are integers such that $x \equiv 3 (mod12)$ and $y \equiv 7(mod18)$, then $x + y \equiv 4(mod6)$

Work in the modulus that represents the least common multiple of $(12)$ and $(18)$, namely $(36)$.

$x \equiv 3\pmod{12} \implies ~\exists ~r \in \Bbb{Z}~$ such that $x = 12r + 3.$

$y \equiv 7\pmod{18} \implies ~\exists ~s \in \Bbb{Z}~$ such that $y = 18s + 7.$

Therefore, $$3x = 36r + 9 ~~~\text{and} ~~~2y = 36s + 14\tag1$$.

The original assertion, $x + y...$ involves an expression where the coefficients applied to $x$ and $y$ are equal.

Therefore, the equations in (1) above must be adjusted so that the coefficients applied to $x$ and $y$ are equal.

Adjusting,

$$6x = 36(2r) + 18 ~~~\text{and} ~~~6y = 36(3s) + 42\tag2$$.

Therefore, $6(x + y) = 36(2r + 3s) + 60$.

Therefore, $$(x + y) = 6(2r + 3s) + 10 \implies (x + y) \equiv 10 \equiv 4 \pmod{6}.$$

2 Answers2