Question

Find the smallest positive integer value of n for which
(1/a)+(1/b) = (1/n)
has at least three solutions (a, b) in integers with
a ≥ b > 0.
A. 3
B. 4
C. 5
D. 6
E. 8

Answer 1

I know that it works for B, D, and E. How would I check A?

Answer 2

Well, I don't have a clever way to do it, but perhaps brute force? Since \(n=3\), we know that \(a,b>3\). Then\[\frac{1}{4}+\frac{1}{b}=\frac{1}{3}\implies b=12.\]\[\frac{1}{5}+\frac{1}{b}=\frac{1}{3}\implies \text{ no solution}.\]\[\frac{1}{6}+\frac{1}{b}=\frac13\implies b=6.\]\[\frac{1}{7}+\frac{1}{b}=\frac13\implies \text{no solution}.\]In fact, for \(7\le a\le11\), there are no solutions. Then, once \(a=12\), this is the same as the first case, so we don't count it. And if \(a\ge 13\), then we also know that \(b\ge 13\) or we would have already seen the solution. But \(1/13+1/13<1/3\). So this will never happen.

Thus, it isn't possible with \(n=3\).

Answer 3

Cool, thanks.

Answer 4

I suppose I should also point out that we also have the condition \(a\ge b>0\), so you may need to reorder the fractions in some of those cases.