Given integers, $s= 2a+3b, t=9a+5b$, if $17\mid s$, prove $17 \mid t$.

Question

Eliminate either $a$ or $b$ to get a value of other variable, i.e. $b,a$ respectively with coefficient as $17$. Say, if eliminate $a$, then get: $$17b = 9s - 2t -(i)$$ else get $$17a = 3t-5s -(ii)$$ Now, using the fact that there are three terms; can use the linear combination property that the set of divisors (or Power set of divisors) are the same for all three terms, say for $17b, 9s, 2t$ in (i).
So in (i), if $17\mid s$, then $17\mid 2a+3b$.
How to take this to conclude that $17\mid t$ also is not clear.

Answer 1

Another perhaps simpler approach. We have that:

\begin{align} s = 2a+3b &\implies a = \frac{s-3b}{2} \tag{1}\\ t & = 9a+5b \tag{2}\\ Sub &~(1) (2) \\ t& = 9(\frac{s-3b}{2})+5b \\ 2t& = 9s-27b+10b \\ 2t& = 9s-17b \end{align}

Clearly $17$ divides $RHS$ because $17|s$ and $17 |17b$. Hence $17$ must divide $LHS$. And since $2$ and $17$ are primes, $17|t$.

Answer 2

If $17\mid2a+3b$, then $17\mid13\times(2a+3b)$. But this means that $17\mid9a+5b$, since$$13\times2\equiv9\pmod{17}\text{ and }13\times3\equiv5\pmod{17}.$$

Answer 3

Suppose that $17$ divides $s$, $2t=18a+10b=9(2a+3b)-17b$. Since $17$ divides $2a+3b$, it divides $2t$. This implies that $17$ divides $t$.

Answer 4

Notice that $$ 3t-5s=17a $$ Multiply by $6$ to get $$ 18t-30s=6\cdot17a $$ Add $30s-17t$ to both sides $$ t=17(6a-t)+30s $$

Answer 5

Note that $$4s+t=17a+17b$$ and all terms apart from $t$ are known to be divisible by $17$