Question

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

Answer 1

Step $1$: Identify the series pattern.

We know that if a number is divisible by $a$ and $b$ both then it should be divisible by LCM of $(a,b)$.

Thus, the two digit number should be divisible by LCM of $\left(2,3\right)=6.$

Two digit numbers divisible by both $2$and $3$ are : $12,18,24..............96$

We can see that it's an A.P. with $a=12$, $a_n=96$ and $d=6$.

Step $2$: Find the two-digit total number of terms in A.P.

$n^{th}$ term of an AP is given by :

$a_n=a+(n-1)d$

Substitute the values.

$$\begin{gathered} 96=12+(n-1)6 \\ \Rightarrow 96-12=(n-1)6 \\ \Rightarrow 84=(n-1)6 \\ \Rightarrow \frac{84}{6}=(n-1) \\ \Rightarrow 14=(n-1) \\ \Rightarrow n=15 \end{gathered}$$

Hence, there are $15$ two-digit numbers that are divided by both $2$and $3$.