# 2005 JBMO Problems/Problem 4

Let (a, b, c) = k Then a = pk

b = qk

c = rk

100a + 10b + c = abc(a + b + c)

100p + 10q + r = pqr(p + q + r)k^3

We see that if any of (a, b, c) is 0, all the terms are 0, and such (a, b, c) are nonzero

We will show that we can find (m, n, t) such that m, n, and t are factors of a, b, and c respectively, with (m, n, p) being pairwise relatively prime.

Let (p, q) = c with p = cs and q = cy

Then

c(100s + 10y) + r = (c^2)syr(cs + cu + r)k^3

We see that c is also a factor of r.

We get similar results for (p, r) = c and (q, r) = c

The terms cancel, and thus we get that

For some (m, n, t) which are respectively factors of (a, b, c)

100m + 10n + t = mnt(m + n + t)j^3

With 0 < m, n, t, j < 10

And (m, n, t) being pairwise relatively prime

The computation is left to the reader

-igetit