Question

How many pairs of positive integers $x,y$ exist such that $\text{HCF}(x,y)+\text{LCM}(x,y)=91$?

• A. $6$
• B. $8$
• C. $7$
• D. $9$

Answer 1

The correct option is B $8$

Let $x=h\times a,y=h\times b$
where $a$ and $b$ are co-prime.
So, HCF of $(x,y)$ is $h$ and LCM of $(x,y)$ is $h\times a\times b$
$\Rightarrow h+h\times a\times b=91$
$\Rightarrow h(ab+1)=91$
Now, $91$ can be written as $1\times 91$ or $7\times 13$

Case $:1$
$h=1$ and $ab+1=91$
$\Rightarrow h=1$ and $ab=90$
There are $4$ pairs of numbers like this : $(1,90),(2,45),(5,18)$ and $(9,10)$

Case $:2$
$h=7$ and $ab+1=13$
$\Rightarrow ab=12$ $\Rightarrow 1\times 12$ or $4\times 3$
Therefore, the pairs of numbers are $(7,84)$ or $(28,21)$

Case $:3$
$h=13$ and $ab+1=7$
$\Rightarrow ab=6\Rightarrow 1\times 6$ or $2\times 3$
Therefore, the pairs of numbers are $(13,78)$ or $(26,39)$

Hence, we have total $8$ possible pairs : $(1,90),(2,45),(5,18),(9,10),(7,84),(28,21),(13,78)$ and $(26,39).$