1976 IMO #4

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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. \(\square\)


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