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Problem of the Week
Problem D and Solution
Using Leftovers


A three-digit positive integer \(n\) has the property that when \(2024\) is divided by \(n\), the remainder is \(4\). What is the sum of all such three-digit positive integers \(n\)?


Let \(p\) be the quotient when \(2024\) is divided by \(n\). Since the remainder is \(4\), it follows that \(np+4=2024\). Thus, \(np=2020\).

Using the prime factorization of \(2020\) we obtain \(2020=2 \times 2 \times 5 \times 101\). From this we can determine all the possible pairs of positive integers that multiply to \(2020\). These are summarized below. \[1 \times 2020, \quad 2 \times 1010, \quad 4 \times 505, \quad 5 \times 404, \quad 10 \times 202, \quad 20 \times 101\] Since \(n\) is a three-digit positive integer, it follows that the only possible values for \(n\) are \(101\), \(202\), \(404\), or \(505\). The sum of these is \(101+202+404+505=1212\).