How many two-digit numbers are divisible by both 2 and 4?

To find the number of 2-digit numbers divisible by both 2 and 4.

A number divisible by both 2 and 4 is divisible by 4.

For an arithmetic sequence, the n-th term is given by:

The sequence of 2-digit numbers divisible by 4 forms an arithmetic sequence (AP) where:

The first term a = 12 (smallest 2-digit number divisible by 4).

The last term l = 96 (largest 2-digit number divisible by 4).

The common difference d = 4.

Now, substituting the values;

96 = 12 + (n – 1) × 4

96 – 12 = (n – 1) × 4

$$\frac{84}{4}$$​ = (n – 1)

n – 1 = 21

n = 22

Thus, there are 22 two-digit numbers divisible by both 2 and 4.