I agree .

If you want a mathematical justification read on. We need a few facts toprovethat no matter how hard we try that is the best possible answer.

First, if are integers (whole numbers) then . You might say the RHS is exponents and LHS is just addition so it makes sense that RHS is much larger than LHS so maybe in fact we should have instead of . But that is not completely true. For small values we actually have equality.

Second, any integer can be written in the form where are integers. Do you see why this is true? See if you can figure this out.

Third, any representation that has amongs its summands (for example ) cannot be the maximum value. The reason is the follows. Say is a representation then its product is . While is also a representation and its product is . Which one is larger? The second one definitely because in the first product the first numbers are the same as in the second product while in the second product the last number, , is larger than the last number, , in the first product.

Now we are ready to state the main result.

Theorem:The maximum product for the representation is achieved where all the summands are 's and 's.

Proof:Say that is a representation. To be maxed we require that each term , look at the third fact we are using. Then each summand can be written as a sum of twos and threes (because each is ) by the second fact. Meaning and ... and . Now by the first fact we are using. But the point is we decomposed each into 's and 's. Meaning , in fact the number of times appears is times and appears times. Which means this new representation is at least as big as the previous one. Thus, any representation which is maxed must be made out of 's and 's.

Finally we can solve the problem. Say that you have twos and threes then and their product is . Thus, we want to maximize subject to . But this is simple. The only possibilities are: and its respective sums are: . Thus, the max product is .