Question

Divide $81$ into two parts so that one may be a multiple of $8$ the other of $5$.

• A. $56,25$
• B. $16,65$
• C. Neither A nor B
• D. Both A and B

Answer 1

Answer: (D)

Both A and B

Let the two parts be $8x$ and $5y$ where $x$ and $y$ are positive integers.

$8x+5y=81$ ......(i)

Thus $\frac{8x}{5}+y=\frac{81}{5}$

$\Rightarrow \frac{3x}{5}+x+y=16+\frac{1}{5}$

$\Rightarrow \frac{3x-1}{5}+x+y=16$

As $x$ and $y$ are positive integers

$\Rightarrow \frac{3x-1}{5}=$ integer

multiplying by $2$

$\Rightarrow \frac{6x-2}{5}=$ integer

$x+\frac{x-2}{5}=$ integer

$\frac{x-2}{5}=$ integer

Let the integer be $p$

$\Rightarrow \frac{x-2}{5}=p$

$\Rightarrow x=5p+2$ .....(ii)

Substituting $x$ in (i), we have

$8(5p+2)+5y=81$

$\Rightarrow 5y=65-40p$

$\Rightarrow y=13-8p$ ......(ii)

From (iii) we can see that $y<0$ for integer $p>1$ which is not taken in our case.

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

Substituting $p$ in (ii) (iii), we get

$\Rightarrow x=2,7$

Substituting $p$ in (iii), we get

$\Rightarrow y=13,5$

According to our assumption two parts were $8x$ and $5y$.

So, the two parts are $16,65$ and $56,25$.