 # How many two-digit numbers are divisible by both 2 and 3?

To find how many two-digit numbers are divisible by both 2 and 3

### Solution

First let us see how many two digits are divisible by 3

First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 12,15,18,…,99

here

First term (a) = 12

Common difference (d) = 3

Let us consider there are n numbers then

an = 99

a + (n – 1)d = 99

12 + (n – 1)3 = 99

12 + 3n – 3 = 99

n = 29+1

n = 30

∴ Two digit numbers divisible by 3 = 30.

Now let us see how many two digits are divisible by 2

First two digit number divisible by 2 = 10

Last two digit number divisible by 3 = 98

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 10, 12,15,18,…,98

here

First term (a) = 12

Common difference (d) = 2

Let us consider there are n numbers then

an = 98

a + (n – 1)d = 98

12 + (n – 1)2 = 98

12 + 2n – 2 = 98

n=45

∴ Two digit numbers divisible by 2 = 45.

Hence total two-digit numbers divisible by 2 and 3 are = 75