Let $a$ and $b$ be integers. If $17\mid 26a + 39b$, prove that $17\mid 9a + 5b$

Question

Okay so I have this exercise question. I tried looking for solutions online and found some related ones but this case is a bit different. I tried by eliminating $a$ also but couldn't get some reasonable answer. Here's my work so far according to another question. Let $p$ = $26a + 39b$ $q$ = $9a + 5b$, $a$ = $\frac{p-39b}{26}$ Substituting in $q$ gives $26q = 9p - 346b$ Now $9p$ is divisible by $17$ but how do I prove for $346b$?

Answer 1

$17 \mid 26a + 39b$

prove $17 \mid 9a +5b$

One side of the proof $\Rightarrow$ you can work backwards to get $\Leftarrow$ :

$17\mid 26a + 39b \iff 9a +17a + 5b +34b \iff(9a+5b) + 17(a + 2b)$

If $17\mid 26a + 39b$ then $17\mid 9a+5b + 17(a+2b)$.

Since $17\mid 17(a+2b)$ and $17\mid 9a+5b + 17(a+2b)$,

$17\mid 9a+5b$ because $17\mid 9a+5b + 17(a+2b)$.

Answer 2

$26a + 39b = (9a +5b) + 17(a + 2b)$ so.....

Answer 3

$$ 13(2a+3b)=17m\Rightarrow 2a+3b=17m'\\ 9a+5b=17n $$

then solving for $a,b$

$$ 2a+3b=17m'\\ 9a+5b=17n $$

$$ a = 3n-5m'\\ b=9m'-2n $$

hence those are the feasible $a,b$

Answer 4

$$26 a =17a +9a $$

$$ 39 b = 34b +5b $$

Upon adding together you get

$$26 a +39 b = 17( a+2b) + 9a +5b$$

$$ 17| ( 26 a +39 b) \implies 17| 9a +5b$$

Answer 5

Given $17 \mid 26a + 39b$ and $a,b\in Z$, note that: $$\frac{26a+39b}{17}-a=\frac{9a+39b}{17}; \\ \frac{9a+39b}{17}-2b=\frac{9a+5b}{17}.$$ Alternatively: $$\frac{26a+39b}{17}=\frac{17a+9a+17b+17b+5b}{17}=a+2b+\frac{9a+5b}{17}.$$