# 2000 PMWC Problems/Problem T8

## Problem

There are positive integers , , such that . What is the smallest possible value of ?

## Solution

Given that , , and are positive integers, it is obvious that none of them can be 1, because that would immediately make the quantity too large.

What about having one of them equal ? We can actually prove through casework that at least one of the integers must be equal to by considering what would happen if none of , , or were equal to . If this was true, then the smallest , , or could be would be . If all three were equal to , then we would have , which is too large. However, the next largest value for the sum would be , which is too small. Thus there are no possibilities that have no s, meaning we can set any one of the variables , , or equal to and continue solving.

Let's say we choose to equal . Then we have . Obviously we can't have another , but what if we had a ? It turns out that we can prove that we must have a in much the same way we proved that there must be a . If we assume that there is no , the largest value of would be , which is too large. Just as last time, the next largest possibility, , is too small. Thus, there are no possibilities that have no s, meaning one of the remaining variables must be . If we randomly select to equal , we are left with the equation , which means . This inequality shows that can be equal to or . The problem asks for the least possible value of , so we say that .

To conclude, we have , which is indeed greater than and less than . This configuration, is one of two that satisfies the given conditions, the other being . Of these two solutions, has the lower sum, meaning our answer is

## See Also

Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems