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

 How many two-digits numbers are divisible by 3 Solution; The two-digit numbers divisible by 3 are 12, 15, 18, …, 99. Clearly, these number are in AP. Here, a = 12 and d =15 —12 = 3 Let this AP contains n terms. Then, an = 99 12 + (n – 1) x 3=99         [an=a+(n —1)cl] 3n-F9=99 3n = 99-9 = 90 n = 30 Hence, there are 30 two-digit numbers divisible by 3. Unacademy is India’s largest online learning platform. Download our apps to start learning Call us and we will answer all your questions about learning on Unacademy AnswerVerifiedHint: 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.