Question

Find the sum of all the three digits natural numbers which on division by 7 leaves remainder 3?

Answer 1

Hint: We start solving the problem by finding the remainder obtained when 100 is divided with 7. We then add or subtract the required number to 100 in order to get the first three digits natural number satisfying the given number. Similarly, we divide 999 and follow the same process as did for the first three digits number to find the last three digits number. We then find the total number of terms present in the obtained series using ${{T}_{n}}=a+\left( n-1 \right)d$. This will be used to find the sum of series using the formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$.

Complete step by step answer:
According to the problem, we need to find the sum of all the three digits natural numbers which on division by 7 leaves remainder 3.
We know that the three digits natural numbers start with 100 and end with 999.
Let us divide 100 with 7 and find what is the remainder in this division process.
$\begin{align}
  & \left. 7 \right)100\left( 14 \right. \\
 & \underline{\text{ 7 }} \\
 & \text{ }30 \\
 & \underline{\text{ }28\text{ }} \\
 & \text{ }2 \\
\end{align}$.
We get the remainder 2 when we divide 100 with 7 and we need the three digits natural numbers which give remainder 3 on division with 7. So, let us add 1 to 100 in order to get the first three digits number which give remainder 3 on division with 7.
So, the first three digits number satisfying our requirement is 101. We then add 7 to get the next terms satisfying our terms.
So, we get the terms as 101, 107, 114, 121,……
Let us find the last term that gives remainder 3 on division with 7.
Let us divide 999 with 7 and find what is the remainder in this division process.
$\begin{align}
  & \left. 7 \right)999\left( 142 \right. \\
 & \underline{\text{ 7 }} \\
 & \text{ }29 \\
 & \underline{\text{ }28\text{ }} \\
 & \text{ }19 \\
 & \underline{\text{ }14\text{ }} \\
 & \text{ }5 \\
\end{align}$.
We get the remainder 5 when we divide 999 with 7 and we need the three digits natural numbers which give remainder 3 on division with 7. So, let us subtract 2 to 999 in order to get the last three digits number which give remainder 3 on division with 7.
So, the last three digits number satisfying our requirement is 997.
Now, let us write the series of numbers that gives remainder 3 on division with 7.
So, we get the series as 101, 107, 114, 121,……, 997.
We can see that the given series resembles arithmetic progression with first term 101, common difference 7 and last terms 997.
We know that the sum of first n terms of A.P is defined as ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$, where n is total no. of terms in A.P, a is first term and l is last term.
So, let us find the total number of terms present in the series as 101, 107, 114, 121,……, 997.
We know that the general term of the series is defined as ${{T}_{n}}=a+\left( n-1 \right)d$, where d is a common difference.
So, we have $997=101+\left( n-1 \right)7$.
$\Rightarrow 997=101+7n-7$.
$\Rightarrow 997=94+7n$.
$\Rightarrow 903=7n$.
$\Rightarrow n=129$.
So, we have found that the series has 129 terms.
Let us find the sum of all the terms in the series 101, 107, 114, 121,……, 997.
We get $S=\dfrac{129}{2}\times \left( 101+997 \right)$.
$\Rightarrow S=\dfrac{129}{2}\times \left( 1098 \right)$.
$\Rightarrow S=129\times 549$.
$\Rightarrow S=70821$.

So, we have found the sum of all the three digits natural numbers which on division by 7 leaves remainder 3 as 70821.

Note: Whenever we get this type of problem, we should start dividing the small number to check whether it is divisible with the given number or not. We should not make calculation mistakes while solving this problem. We can also use ${{S}_{n}}=\dfrac{n}{2}\times \left( 2a+\left( n-1 \right)d \right)$ to find the sum of the terms in obtained series. Similarly, we can expect problems to find the sum of all the three digits natural numbers which is divisible by 11.