Question

Solve in positive integers:
$13x+11y=414$.

• A. $(9,27)$
• B. $(6,13)$
• C. $(31,1)$
• D. $(5,33)$

Answer 1

Answer: (A), (C)

The correct options are
A $(9,27)$
C $(31,1)$

Given, $13x+11y=414$

$\Rightarrow \frac{13x}{11}+y=\frac{414}{11}$

$\Rightarrow y+x+\frac{2x}{11}=37+\frac{7}{11}$

$\Rightarrow y+x+\frac{2x-7}{11}=37$

As we are solving for positive integers, so $x$ and $y$ are both integers.

$\Rightarrow \frac{2x-7}{11}=$integer

Multiplying by $6$, we get

$\Rightarrow \frac{12x-42}{11}=$ integer

$x-3+\frac{x-9}{11}=$ integer

$\Rightarrow \frac{x-9}{11}=$ integer

Let the integer be $p$

$\frac{x-9}{11}=px=11p+9$ .........(ii)

Substituting $x$ in (i), we get

$13(11p+9)+11y=414\Rightarrow 11y=297-141p\Rightarrow y=27-13p$ .......(iii)

From (ii) we see that $x<0$ for $p<0$ and from (iii) we see that $y<0$ for $p>1$ which is not possible as we are solving for only positive integers.

So, the values of $p$ can be $0,1,2$

Substituting $p$ in (ii), we get

$\Rightarrow x=9,20,31$

Substituting $p$ in (iii)

$\Rightarrow y=27,14,1$

So, the complete solution set of positive integers is

$\{\begin{array}{c}x=9,20,31\\ y=27,14,1\end{array}$