Question

Solve in positive integers:
$23x+25y=915$

• A. $x=15,y=6$
• B. $x=30,y=9$
• C. $x=5,y=32$
• D. $x=10,y=8$

Answer 1

Answer: (B), (C)

The correct options are
B $x=5,y=32$
C $x=30,y=9$

$23x+25y=915$ ......(i)

$\Rightarrow x+\frac{25y}{23}=\frac{915}{23}\Rightarrow x+y+\frac{2y}{3}=39+\frac{18}{23}\Rightarrow x+y+\frac{2y-18}{3}=39$

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

$\Rightarrow \frac{2y-18}{23}=$integer

Multiplying by $12$, we get

$\Rightarrow \frac{24y-216}{23}=$ integer

$\Rightarrow y-9+\frac{y-9}{23}=$ integer

$\Rightarrow \frac{y-9}{23}=$ integer

Let the integer be $p$.

$y-9=23p\Rightarrow y=9+23p$ ......(ii)

Substituting $y$ in (i), we get
$23x+25(9+23p)=915\Rightarrow 23x=690-575p\Rightarrow x=30-25p$

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

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

substituting $p$ in (iii), we get

$\Rightarrow y=9,32$

Substituting $p$ in (iii),

$\Rightarrow x=30,5$

So, the complete solution set is

$\{\begin{array}{c}x=30,5\\ y=9,32\end{array}$