Question

Solve in positive integers:
$3x+8y=103$.

• A. $(29,2)$
• B. $(32,3)$
• C. $(25,4)$
• D. $(24,9)$

Answer 1

The correct option is A $(29,2)$

Given equation is $3x+8y=103$ ......(i)$

Dividing by $3$, we get

$x+\frac{8y}{3}=\frac{103}{3}\Rightarrow x+2y+\frac{2y}{3}=34+\frac{1}{3}\Rightarrow x+2y+\frac{2y-1}{3}=34$

We have to solve for positive integers, so $x$ and $y$ are integers

which $\frac{2y-1}{3}=$ integer

On multiplying by $2$, we get

$\frac{4y-2}{3}=$ integer

$y+\frac{y-2}{3}=$ integer

$\Rightarrow \frac{y-2}{3}=$ integer

Let the integer be $p$

$\frac{y-2}{3}=p\Rightarrow y=3p+2$ .......(ii)

Substituting $y$ in $\left(i\right)$

$3x+8(3p+2)=1033x=87-24px=29-8p$ ......(iii)

From (ii) we see that $y<0$ for integers $<0$ and from (iii) we see that $x<0$ for $p>3$ which is not possible as we are solving for positive integers only.

So, $p$ can be equal to $0,1,2,3$.

Substituting $p$ in (ii), we get

$\Rightarrow y=2,5,8,11$

Substituting $p$ in $\left(iii\right)$

$\Rightarrow x=29,19,13,5$

So, the complete solution set of positive integers is

$\{\begin{array}{c}x=29,21,13,5\\ y=2,5,8,11\end{array}$