#
Problem
of the Week

Problem
D and Solution

Using
Leftovers

## Problem

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 \(n\)?

## Solution

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 \(101+202+404+505=1212\).