# How many positive 3-digit numbers are multiples of 20, but not of 55?

## ~ Question from AoPS ~ Background Information Subject: Prealgebra Focus/Topic: Counting and Probability (I think) *Review Problem

41

#### Explanation:

To be a multiple of 20, a natural number would have to be multiplied by 20 to get to it. For instance, 40 is a multiple of 20:

$20 \times 2 = 40$

and 66 isn't:

$20 \times 3 = 60$

$66$

$20 \times 4 = 80$

And the same goes for 55 and its multiples.

So let's first look at the number of multiples of 20 are 3-digit numbers. The lowest 3-digit number that is a multiple of 20 is 100:

$20 \times 5 = 100$

and the largest is 980:

$20 \times 49 = 980$

That's $49 - 5 + 1 = 45$ multiples (we add 1 to include the first multiple).

For a number to be both a multiple of 20 and 55, we can look first at the LCM by doing a prime factorization:

$20 = 2 \times 2 \times 5$
$55 = 5 \times 11$

And so the LCM is:

$2 \times 2 \times 5 \times 11 = 220$

Every multiple after that will be a multiple of 220:

$220 \times 2 = 440$
$220 \times 3 = 660$
$220 \times 4 = 880$

notice that the next multiple will be a 4-digit number and so we're not interested in it for this problem.

And so there are 4 numbers that are multiples of both 20 and 55. We want to know how are multiples of 20 and not 55, so we subtract this result from our multiples of 20 result:

$45 - 4 = 41$