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2006 USAMO Problems/Problem 4

Problem

Find all positive integers $n$ such that there are $k\ge 2$ positive rational numbers $a_1,a_2,\ldots a_k$ satisfying $a_1+a_2+...+a_k = a_1 \cdot a_2 \cdot \cdots a_k = n$.

Solution

First, consider composite numbers. We can then factor $n$ into $p_1*p_2.$ It is easy to see that $p_1+p_2\le n$, and thus, we can add $(n-p_1-p_2)$ 1s in order to achieve a sum and product of $n$. For $p_1+p_2=n$, which is only possible in one case, $n=4$, we consider $p_1=p_2=2$.

Secondly, let $n$ be a prime. Then we can find the following procedure: Let $a_1=\frac{n}{2}, a_2=4, a_3=\frac{1}{2}$ and let the rest of the $a_k$ be 1. The only numbers we now need to check are those such that $\frac{n}{2}+4+\frac{1}{2}>n\Longrightarrow n<9$. Thus, we need to check for $n=1,2,3,5,7$. One is included because it is neither prime nor composite.

For $n=1$, consider $a_1a_2\hdots a_k=1$. Then by AM-GM, $a_1+a_2+\hdots+a_k\ge k\sqrt[k]{1}>1$ for $k\ge 2$. Thus, $n=1$ is impossible.

If $n=2$, once again consider $a_1a_2\hdots a_k=2$. Similar to the above, $a_1+a_2+\hdots\ge k\sqrt[k]{2}>2$ for $k\ge 2$ since $\sqrt[k]{2}>1$ and $k>2$. Obviously, $n=2$ is then impossible.

If $n=3$, let $a_1a_2\hdots a_k=3$. Again, $a_1+a_2+\hdots\ge k\sqrt[k]{3}>3$. This is obvious for $k\ge 3$. Now consider $k=2$. Then $2*\sqrt{3}\approx 3.4$ is obviously greater than $3$. Thus, $n=3$ is impossible.

If $n=5$, proceed as above and consider $k=2$. Then $a_1+a_2=5$ and $a_1a_2=5$. However, we then come to the quadratic $a_1^2-5a_1+5=0 \Longrightarrow a_1=\frac{5\pm\sqrt{5}}{2}$, which is not rational. For $k=3$ and $k=4$ we note that $\sqrt[3]{5}>\frac{5}{3}$ and $\sqrt[4]{5}>\frac{5}{4}$. This is trivial to prove. If $k\ge 5$, it is obviously impossible, and thus $n=5$ does not work.

The last case, where $n=7$, is possible using the following three numbers. $a_1=\frac{9}{2}, a_2=\frac{4}{3}, a_3=\frac{7}{6}$ shows that $n=7$ is possible.

Hence, $n$ can be any positive integer greater than $3$ with the exclusion of $5$.

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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