# 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 $$10$$.