A number theory problem by Maruf Ahmed

Let the two numbers be $a$ and $b$ , where $a < b$ and hence $a = \frac 23 b$ . Using the c onnection between LCM and GCD , we have:

$\begin{aligned} \text{lcm}(a,b) \times \gcd(a, b) & = ab \\ \implies ab & = 60 \times 10 = 600 \\ \frac 32 a^2 & = 600 \\ a^2 & = 400 \\ \implies a & = \boxed{20} \end{aligned}$

We can make use of the relation $\gcd(x,y) \times \text{lcm}(x,y) = x \times y$ to solve this. Since we're interested in the smaller of the two numbers, put $y=\frac32 x$ , and substitute in to find $20 \times 60 = x \times \frac32 x$ , which simplifies to $x^2=400$ ; hence $x=\boxed{20}$ and $y=30$ (ignoring the negative root).

It's easy to prove the above relation by considering the number of times each prime factor of $x$ or $y$ occurs in each of $\gcd(x,y)$ and $\text{lcm}(x,y)$ .