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

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

ddcamp · Answer

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

kinggeorge · Answer

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 $n=3$.

ddcamp · Answer

Cool, thanks.

kinggeorge · Answer

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.