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**Hint:**Here, we will first find the total number of two-digit numbers. Then we will write down a few numbers that are divisible by both 2 and 3 and try to find the pattern or relation between each number. Then observing the pattern, we will apply the formula of \[{n^{th}}\] term of an AP and simplify further to get the required answer.

**Complete step by step solution:**

We know that two-digit numbers lie between 10 and 99.

First-term that is divisible by 3 is 12 which is also divided by 2 then we have 18 which is divisible by 2 and 3 both then we have 24 which is again divisible by 2 and 3.

So, the numbers divisible by both 2 and 3 are \[12,18,24,......96\].

As we can see that the above relation is an A.P with first term as 12, common difference as 3 and \[{n^{th}}\] term as 96.

Substituting \[a = 12,d = 6\] and \[{T_n} = 96\]in the formula \[{n^{th}}\] term of AP \[{T_n} = a + d\left( {n - 1} \right)\], we get

\[96 = 12 + 6\left( {n - 1} \right)\]

Multiplying the terms, we get

\[ \Rightarrow 96 = 12 + 6n - 6\]

Rewriting the equation, we get

\[ \Rightarrow 6n = 96 - 12 + 6\]

Adding and subtracting the like terms, we get

\[ \Rightarrow 6n = 90\]

Dividing both side by 6, we get

\[ \Rightarrow n = 15\]

**Therefore, we can conclude that there are 15 two-digits numbers that are divisible by both 2 and 3.**

**Note:**

The divisibility rule is a method by which we can easily find whether a number is divisible by a particular number or not. We can only say that to find numbers divisible by 2 and 3 both we follow divisibility of 6 or we can say if a number is divisible by 2 and 3 it also satisfies the divisibility of digit 6. Here, the numbers that are divisible by both 2 and 3 form an AP series. AP or an arithmetic progression is a series or sequence in which the consecutive numbers differ by a common difference.

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