So, the question is that the sum of HCF and LCM is $96$ and the sum of the numbers is $48$. We need to find the numbers.
Here is my attempt to this question:
Let the numbers be $a$ and $b$ and their LCM and HCF be $l$ and $h$ respectively. So, the 3 equations that we have are
$$a + b = 48$$
$$l + h = 96$$
$$ab = lh$$
Substituting the value of $a$ as $48 – b$ and $l$ as $96 – h$ in $ab = lh$ we get
$$(48 – b)b = (96 – h)h \\ \implies 48b – b^2 = 96h – h^2 \\ \implies 48b – 96h = b^2 – h^2$$
On comparing LHS with RHS we get $b$ as $48$ and $h$ as $96$. However, this would mean that LCM and $a$ are $0$ which is not true as LCM can't be less the HCF or the numbers. Is there some other way of doing it?
Best Answer
$$a + b = 48$$ $$l + h = 96$$ $$ab = lh$$
So $$l = ab/h$$
Let $a = ph$, $b = qh$
Therefore $$ph + qh = 48$$ $$pqh + h = 96$$ Thus $$pqh + h = 2(ph + qh)$$ $$pq + 1 = 2p + 2q$$ $$pq -2p -2q = -1$$ By adding 4 to both sides we can factor the LHS $$pq -2p -2q + 4 = 3$$ $$(p-2)(q-2) = 3$$ 3 is prime, so one of $p-2$ and $q-2$ must be 3 and the other must be 1 (since we're working with positive integers).
WLOG, let $p-2 = 3$ and $q-2 = 1$ Therefore $p=5$ and $q=3$
$p+q=8$ and since $(p + q)h = 48$
$h = 48/8 = 6$
So $a = 5.6 = 30$ and $b = 3.6 = 18$
and $l = LCM(5.6, 3.6) = 5.18 = 30.3 = 90$
Hence $l + h = 96$