Let $HCF (x, y) = h$ then we can say that $x=ha$ and $y=hb$ for some multiple of $h$ i.e. $a$ and $b$ respectively.
Let $LCM(x, y) = k$
We know that :-
$HCF (x, y) \times LCM(x, y) = x \times y$
$\Rightarrow h \times k = x \times y $
$\Rightarrow h \times k = ha \times hb $
$\Rightarrow k = h \times a \times b $
As per question : –
$HCF (x, y) + LCM (x, y) = 91$
$\Rightarrow h + k =91$
$\Rightarrow h + hab =91$
$\Rightarrow h(1+ab) =91$
Case $1$ when $h= 1$ then $ab =90$ :-
We get the solution pair for $\{a,b\} = \{(1,90),(2,45),(3,30),(5,18),(6,15),(9,10)\}$
We get the solution pair for $\{x,y\} = \{ha,hb\} = \{(1,90),(2,45),(3,30),(5,18),(6,15),(9,10)\}$
Case $2$ when $h= 7$ then $ab =12$ :-
We get the solution pair for $\{a,b\} = \{(1,12),(2,6),(3,4)\}$
We get the solution pair for $\{x,y\} = \{ha,hb\} = \{(7,84),(14,42),(21,28)\}$
Case $3$ when $h= 13$ then $ab =6$ :-
We get the solution pair for $\{a,b\} = \{(1,6),(2,3)\}$
We get the solution pair for $\{x,y\} = \{ha,hb\} = \{(13,78),(26,39)\}$
Case $4$ when $h= 91$ then $ab =0$ which is not a valid case for our solution.
So you can see that I have got $11$ pairs of positive integers which satisfy the given condition but this is not the correct answer that has been provided. I am wondering what I have done wrong in this. Can someone please help me on this?
Thanks in advance !
Best Answer
Hint: Note that with $\operatorname{HCF}(x,y) = h$, along with $x = ha$ and $y = hb$, then we also have that $\operatorname{HCF}(a,b) = 1$. However, this requirement is not satisfied in some of your proposed solutions so they can't be included, e.g., for your Case $1$, you've included $(a,b) = (3,30)$ and $(a,b) = (6,15)$, both of which have $\operatorname{HCF}(a,b) = 3$ (note also that $h = 3 \neq 1$ for those $2$ proposed solutions).