How many 2 digit numbers are divisible by 8?
A. 12
B. 11
C. 10
D. 13

Hint: We need to find the numbers of 2 digit numbers that are divisible by 8. For this, we will form an arithmetic progression of two-digit numbers in which the first term will be the first 2 digit number divisible by 8 and the last term will be the last 2 digit number divisible by 8. The AP will contain the multiples of 8 only thus the common difference will be 8. Then we will calculate the number of terms by the formula ${{a}_{n}}=a+\left( n-1 \right)d$ where a is the first term of the AP, n is the number of terms in the AP, ${{a}_{n}}$ is the last term of the AP and d is a common difference.

 Complete step-by-step solution:
Now, we know that the first 2 digit number divisible by 8 is ‘16’.
Therefore, if ‘a’ is the first term of the AP then:
$a=16$
Now, we also know that the last 2 digit number which is divisible by 8 is ‘96’.
Therefore, if $'{{a}_{n}}'$ is the last term of the AP then:
${{a}_{n}}=96$
Now, we will calculate the number of terms in the AP, i.e., the number of 2 digit numbers divisible by 8.
Let ‘n’ be the number of terms in the AP. Thus, by the formula ${{a}_{n}}=a+\left( n-1 \right)d$ we have:
$\begin{align}
  & \Rightarrow 96=16+\left( n-1 \right)8 \\
 & \Rightarrow 96-16=\left( n-1 \right)8 \\
\end{align}$
$\begin{align}
  & \Rightarrow 80=\left( n-1 \right)8 \\
 & \Rightarrow \dfrac{80}{8}=n-1 \\
 & \Rightarrow n-1=10 \\
 & \Rightarrow n=11 \\
\end{align}$
Therefore, the total number of 2 digit numbers divisible by 8 are 11.
Thus, option (B) is the correct option.

Note: If we do not know the last 2 digit number divisible by 8 we can also calculate it by dividing the last 2 digit number, i.e., 99 by 8, and then subtract the remainder from 99. It will give us the last 2 digit number which is divisible by 8.
We can also calculate the number of 2 digit numbers divisible by 8 by the following method:
When we divide 99 by 8, we get 12 as the quotient that means the last 2 digit number divisible by 8 is the ${{12}^{th}}$ multiple of 8, and the first 2 digit number divisible by 8 is the ${{2}^{nd}}$ multiple of 8.
Thus, n= (last factor-first factor) + 1
$\begin{align}
  & \Rightarrow n=\left( 12-2 \right)+1 \\
 & \Rightarrow n=11 \\
\end{align}$