### 1976 IMO #4

Determine the largest number which is the product of positive integers with sum 1976.

The answer is $\boxed{2 \cdot 3^{658}}$.

Note that there cannot be any integers $n>4$ in the maximal product, because we can just replace $n$ by $3,n-3$ to achieve a greater product. Also, we would not want any $1$s in the maximal product as they would not help in increasing the product and would just take up space in the sum. Notice that we can replace any $4$s by two $2$s leaving the product and the sum unchanged. Lastly, we don't want more than two $2$s, because we can just replace three $2$s by two $3$s to obtain a larger product. Therefore, the maximum product must consist of $3$s and zero, one or two $2$s. Since $1976 = 3\cdot 658 + 2$, we have one $2$ and $658$ $3$'s in our maximum product, giving the answer of $2\cdot 3^{658}$, as desired. $\square$