#
Problem
of the Week

Problem
C and Solution

A
Multiple Problem

## Problem

How many integers between \(100\)
and \(2024\) are multiples of both
\(5\) and \(7\), but are not multiples of \(10\)?

## Solution

The integers that are multiples of both \(5\) and \(7\) are the integers that are multiples of
\(35\). Now letâ€™s determine which
multiples of \(35\) are also multiples
of \(10\). Notice that \(1 \times 35 = 35\), which is not a multiple
of \(10\). However, \(2 \times 35 = 70\), which is a multiple of
\(10\). In fact, multiplying \(35\) by any even integer will result in a
multiple of \(10\). This is because
\(35\) is a multiple of \(5\), and all even integers are multiples of
\(2\). So multiplying \(35\) by an even integer will result in an
integer which is a multiple of both \(5\) and \(2\), and thus a multiple of \(10\). So if we are looking for integers
that are multiples of \(35\) but not
multiples of \(10\), then we must
multiply \(35\) by odd integers
only.

The smallest multiple of \(35\)
greater than \(100\) is \(3 \times 35 = 105\). Similarly, the largest
multiple of \(35\) less than \(2024\) is \(57
\times 35 = 1995\). It follows that the number of integers
between \(100\) and \(2024\) that are multiples of both \(5\) and \(7\), but are not multiples of \(10\), is equal to the number of odd
integers between \(3\) and \(57\), inclusive. This is equal to the
number of odd integers between \(1\)
and \(55\), inclusive. We know that
exactly half of the integers between \(1\) and \(54\) are odd, and \(55\) is an odd integer. So in total, there
are \(\frac{54}{2}+1=27+1=28\) odd
integers between \(1\) and \(55\), inclusive.

Thus, there are \(28\) integers
between \(100\) and \(2024\) that are multiples of both \(5\) and

\(7\), but are not multiples of \(10\).