Question

What is the sum of two digit numbers which are divisible by 3 and not 2?

Answer 1

Hint: Here we will proceed first by forming the series of the two digit numbers which are divisible by 3 and then using the formula for the ${{n}^{th}}$ term of A.P, that is, ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ in order to find the number of terms present $\left( n \right)$. And then find the total numbers which are divisible by 2 I this series. And then remove them from the actual series and get the answer.

Complete step by step answer:
As we know, any 2-digit number will have two places (That is unit places and ten places) which will be occupied by the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Clearly the two digits which are divisible by 3 are given by 12, 15, 18, 21, … 99. And the digits which are not divisible by 2 in these digits are all odd digits. So, we will count only odd digits here. As we know that for any A.P with first term as ${{a}_{1}}$ and common difference $d$, the ${{n}^{th}}$ term of A.P,
 ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\ldots \ldots \ldots \left( i \right)$
So, here ${{a}_{n}}=99,{{a}_{1}}=12$ and $d=3$. With the help of these, find the value of $n$.
$\begin{align}
  & {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & \Rightarrow 99=12+\left( n-1 \right)3 \\
 & \Rightarrow 33=4+\left( n-1 \right) \\
 & \Rightarrow 33=3+n \\
\end{align}$
So, the value of $n=33-3=30$.
The sum of all these numbers is given by,
${{S}_{1}}=\dfrac{n}{2}\left( a+l \right)=\dfrac{30}{2}\left( 12+99 \right)=1665$
Now we need to find two digit numbers which are divisible by 6. 12 is the first number and 96 is the last number and the common difference is 6.
$\begin{align}
  & \Rightarrow {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & \Rightarrow 96=12+\left( n-1 \right)6 \\
 & \Rightarrow 16=2+\left( n-1 \right) \\
 & \Rightarrow 16=1+n \\
 & \Rightarrow n=15 \\
\end{align}$
Sum of all these numbers,
${{S}_{1}}=\dfrac{n}{2}\left( a+l \right)=\dfrac{15}{2}\left( 12+96 \right)=810$
So, the sum of all divisible numbers by 3 but not by 2 is 1665 - 810 = 855.

Note: For solving these type of questions you have to first proceed to the total values of the variables according to ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ formula. If all these are found then, it will be very easy to find the rest values for this. And always count the sum by the ${{S}_{1}}=\dfrac{n}{2}\left( a+l \right)$ formula.