Question

Find the sum of all two-digit positive numbers which when divided by 7 yield 2 or 5 as remainder?
(a) 1365
(b) 1256
(c) 1465
(d) 1356

Answer 1

Hint: We start solving the problem by writing the general form of the number that leaves remainder 2 when they are divided by 7. We write all the numbers and take the sum of them. Similarly, we write the general form of the number that leaves remainder 5 when they are divided by 7 and we write all the numbers and take the sum of them. Now, we add both the sums to get the required result.

Complete step-by-step answer:
According to the problem, we need to find the sum of the two-digit positive numbers which when divided by 7 gives 2 or 5 as remainder.
We know that the two-digit positive numbers lie between 10 and 99. We know that the numbers which are divisible by 7 are of the form $7r\left( r\ge 1 \right)$. We know that if a divisor (number) is divided by dividend (another number) and leaves a remainder, then the divisor can be written as $\text{divisor = }\left( \text{dividend}\times \text{quotient} \right)+\text{Remainder}$.
We get the general form of the numbers that when divided by 7 leaves remainder 2 as $7r+2\left( r>1 \right)$. Here $r=1$ is neglected as the number 9 is not present between the numbers 10 and 99.
We get numbers 16, 23, 30,……,93 after substituting the$r=2,3,.....,13$. We can see that these numbers follow A.P (Arithmetic progression) with first term 16 and last term 93. We have a total of 12 terms present here.
We know that the sum of the n-terms of the series in A.P (Arithmetic progression) $a,\left( a+d \right),.......,\left( a+\left( n-1 \right)d \right)$ is $\dfrac{n}{2}\left( a+\left( a+\left( n-1 \right)d \right) \right)=\dfrac{n}{2}\left( \text{first term + last term} \right)$.
Now, we find the sum of the numbers 16, 23, 30,……,93. Let us assume this sum be ${{S}_{1}}$.
We get ${{S}_{1}}=\dfrac{12}{2}\times \left( 16+93 \right)$.
$\Rightarrow {{S}_{1}}=6\times \left( 109 \right)$.
$\Rightarrow {{S}_{1}}=654$ ---(1).
We get the general form of the numbers that when divided by 7 leaves remainder 5 as$7r+5\left( r\ge 1 \right)$.
We get numbers 12, 19, 26,……,96 after substituting the $r=1,2,3,.....,13$. We can see that these numbers follow A.P (Arithmetic progression) with first term 12 and last term 96. We have a total of 13 terms present here.
We know that the sum of the n-terms of the series in A.P (Arithmetic progression) $a,\left( a+d \right),.......,\left( a+\left( n-1 \right)d \right)$ is $\dfrac{n}{2}\left( a+\left( a+\left( n-1 \right)d \right) \right)=\dfrac{n}{2}\left( \text{first term + last term} \right)$.
Now, we find the sum of the numbers 12, 19, 26,……,96. Let us assume this sum be ${{S}_{2}}$.
We get ${{S}_{2}}=\dfrac{13}{2}\times \left( 12+96 \right)$.
$\Rightarrow {{S}_{2}}=\dfrac{13}{2}\times \left( 108 \right)$.
$\Rightarrow {{S}_{2}}=13\times \left( 54 \right)$.
$\Rightarrow {{S}_{2}}=702$ ---(1).
We need to sum the numbers that leave remainder 2 or 5 when they are divided by 7. So, we add ${{S}_{1}}$ and ${{S}_{2}}$ to get the required sum S (assume).
So, $S={{S}_{1}}+{{S}_{2}}$.
$\Rightarrow S=654+702$.
$\Rightarrow S=1356$.
We have found the sum of the numbers that leaves remainder 2 or 5 when they are divided by 7 as 1356.
∴ The sum of the numbers that leaves remainder 2 or 5 when they are divided by 7 is 1356.

So, the correct answer is “Option d”.

Note: We can also solve the problem as follows:
$\Rightarrow {{S}_{1}}=\sum\limits_{2}^{13}{\left( 7r+2 \right)}$.
Here we add and subtract the term that we obtain after substituting $r=1$.
$\Rightarrow {{S}_{1}}=\left( \sum\limits_{1}^{13}{\left( 7r+2 \right)} \right)-\left( 7\left( 1 \right)+2 \right)$.
$\Rightarrow {{S}_{1}}=\left( \sum\limits_{1}^{13}{7r}+\sum\limits_{1}^{13}{2} \right)-\left( 7+2 \right)$.
$\Rightarrow {{S}_{1}}=\left( 7\sum\limits_{1}^{13}{r}+\sum\limits_{1}^{13}{2} \right)-\left( 9 \right)$.
We know that the sum of the n natural numbers is defined as $\dfrac{n\left( n+1 \right)}{2}$ and $\sum\limits_{1}^{n}{a}=an$.
$\Rightarrow {{S}_{1}}=\left( 7\times \dfrac{13\times \left( 13+1 \right)}{2}+\left( 2\times 13 \right) \right)-\left( 9 \right)$.
$\Rightarrow {{S}_{1}}=\left( \dfrac{91\times \left( 14 \right)}{2}+\left( 2\times 13 \right) \right)-\left( 9 \right)$.
$\Rightarrow {{S}_{1}}=\left( \left( 91\times 7 \right)+26 \right)-\left( 9 \right)$.
$\Rightarrow {{S}_{1}}=637+26-9$.
$\Rightarrow {{S}_{1}}=654$.
Similarly,
$\Rightarrow {{S}_{2}}=\sum\limits_{1}^{13}{\left( 7r+5 \right)}$.
\[\Rightarrow {{S}_{2}}=\left( \sum\limits_{1}^{13}{7r}+\sum\limits_{1}^{13}{5} \right)\].
$\Rightarrow {{S}_{2}}=\left( 7\sum\limits_{1}^{13}{r}+\sum\limits_{1}^{13}{5} \right)$.
We know that the sum of the n natural numbers is defined as $\dfrac{n\left( n+1 \right)}{2}$ and $\sum\limits_{1}^{n}{a}=an$.
$\Rightarrow {{S}_{2}}=\left( 7\times \dfrac{13\times \left( 13+1 \right)}{2}+\left( 5\times 13 \right) \right)$.
$\Rightarrow {{S}_{2}}=\left( \dfrac{91\times \left( 14 \right)}{2}+65 \right)$.
$\Rightarrow {{S}_{2}}=\left( \left( 91\times 7 \right)+65 \right)$.
$\Rightarrow {{S}_{2}}=637+65$.
$\Rightarrow {{S}_{2}}=702$.
Now, we have the required sum $S={{S}_{1}}+{{S}_{2}}$.
$\Rightarrow S=654+702$.
$\Rightarrow S=1356$.