Brenda,

The easiest way is to look at the canonical representations of n and s as a product of primes (generalize the content of my note on the MathCentral website), but if you're not familiar with that we can do it as follows.

I'll use (n,s) for the gcd of n and s and use [n,s] for the lcm of n and s.

Write n = (n,s)r and s = (n,s)t; then (t,r)=1 and ns = (n,s)r(n,s)t. Thus ns/(n,s) = r(n,s)t which is clearly a multiple of both n and s.

Suppose there is a smaller integer, m = [n,s], which both n and s divide, then m = un and m = vs for some integers u and v. Write (n,s) = nx + sy for some integers x and y (this can be done as a consequence of the Euclidean Algorithm) then m/(ns/(n,s)) = m(n,s)/ns = (mnx+msy)/ns = (vsnx+unsy)/ns = vx + sy, an integer, thus m is divisible by ns/(n,s) and m ≥ ns/(n,s), a contradiction.

Penny.