Question

The sum of integers lying between 1 and 100 and are divisible by 3 or 5 or 7 is:
(a) 2838
(b) 3468
(c) 2738
(d) 3368

Answer 1

Hint: First of all find the sum of integers which are divisible by 3 then find the sum of integers which are divisible by 5 and also find the sum of integers which are divisible by 7. Now, add the sum of answers of the integers which are divisible by 3, 5, and 7 that we have just calculated. After that find the sum of integers which are divisible by 15 and the sum of the integers which are divisible by 35 and the sum of integers which are divisible by 21 then add the results of these summations of integers which are divisible by 15, 35, and 21. Now, subtract the result of integers which are divisible by 15, 35, and 21 from the result of the summation of integers which are divisible by 3, 5, and 7.

Complete step-by-step solution:
We are going to write the integers which are divisible by 3 lying between 1 and 100 (including 1 and 100).
3, 6, 9, 12, 15……..99
As you can see that the above series is forming an A.P. in which the first term is 3 and the common difference is 3 so the number of terms we can find as follows:
We know the formula for the general term of an A.P. as:
${{T}_{n}}=a+\left( n-1 \right)d$
In the above formula, “a” is the first term and “d” is the common difference of an A.P.
Now, to find the number of terms we are going to put 99 in place of ${{T}_{n}}$ and 3 in place of “a” and 3 in place of “d”.
$\begin{align}
  & 99=3+\left( n-1 \right)3 \\
 & \Rightarrow 99=3+3n-3 \\
\end{align}$
$\begin{align}
  & \Rightarrow 99=3n \\
 & \Rightarrow \dfrac{99}{3}=n \\
 & \Rightarrow n=33 \\
\end{align}$
Now, we are going to calculate the sum of 33 terms as follows:
We know that the sum of n terms which are in A.P. is equal to:
${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
In the above formula, “n” is the total number of terms, “a” is the first term and “l” is the last term of an A.P.
Now, substituting n as 33, “a” as 3 and “l” as 99 in the above summation formula we get,
$\dfrac{33}{2}\left( 3+99 \right)$
$\begin{align}
  & =\dfrac{33}{2}\left( 102 \right) \\
 & =33\left( 51 \right)=1683 \\
\end{align}$
Similarly, we can find the sum of the integers which are divisible by 5 and 7.
Integers between 1 and 100 (including 1 and 100) which are divisible by 5 are:
5, 10, 15, 20, 25…….100
The number of terms is 20.
And sum of the 20 terms is calculated as follows:
$\begin{align}
  & \dfrac{20}{2}\left( 5+100 \right) \\
 & =10\left( 105 \right)=1050 \\
\end{align}$
Integers between 1 and 100 (including 1 and 100) which are divisible by 7 are:
7, 14, 21, 28, 35……..98
The calculation of finding the number of terms is the same as we have shown for the integers which are divisible by 3 so the number of terms in the above series is 14.
And the sum of the 14 terms is calculated as follows:
$\begin{align}
  & \dfrac{14}{2}\left( 7+98 \right) \\
 & =7\left( 105 \right)=735 \\
\end{align}$
Now, adding the result of the sum of integers which are divisible by 3, 5, and 7 we get,
$\begin{align}
  & 1683+1050+735 \\
 & =3468 \\
\end{align}$
Now, as you can see that the integers that we have found above which are divisible by 3 and the integers which are divisible by 5 have some integers in common like 15 is divisible by 3 and 15 is also divisible by 5 so basically, we have counted 15 twice so we have to subtract the integers which are common in the two sets. Similarly, there are integers which are common in the sets of integers which are divisible by 5 and divisible by 7 so that also we have to subtract and also the common integers in the sets in which some integers are divisible by 3 and at the same time they are divisible by 7 also. To mitigate this problem, we are going to find the integers which are divisible by 15, 35, and 21 and then add them and further will subtract them from the total addition of integers which are divisible by 3, 5, or 7.
We are writing the integers which are divisible by 15.
15, 30, 45……. 90
The number of terms in the above series is equal to 6.
The sum of these 6 terms is equal to:
$\begin{align}
  & \dfrac{6}{2}\left( 15+90 \right) \\
 & =3\left( 105 \right)=315 \\
\end{align}$
Now, the integers which are divisible by 35 are:
35, 70
Addition of these integers is equal to:
$\begin{align}
  & 35+70 \\
 & =105 \\
\end{align}$
The integers which are divisible by 21 are:
21, 42, 63, 84
Adding the above integers we get,
$\begin{align}
  & 21+42+63+84 \\
 & =210 \\
\end{align}$
Now, adding the result of sum of integers which are divisible by 15, 35 and 21 we get,
$\begin{align}
  & 315+105+210 \\
 & =630 \\
\end{align}$
Subtracting the result of sum of integers which are divisible by 15, 35 and 21 from the result of sum of integers which are divisible by 3, 5 and 7 we get,
$\begin{align}
  & 3468-630 \\
 & =2838 \\
\end{align}$
From the above solution, we got the sum of integers lying between 1 and 100 and are divisible by 3 or 5 or 7 as 2838.
Hence, the correct option is (a).

Note: The mistake that could happen in the above problem is that you forget to subtract the sum of integers which are divisible by 15, 35, and 21 from the sum of integers which are divisible by 3, 5, and 7. An interesting point is that you will find the sum of integers, which are divisible by 3, 5, and 7 in one of the options. In the above solution, we found the sum of integers which are divisible by 3, 5, and 7 as 3468 and you can see the option (b) is also corresponding to 3468 so here, the chances of making mistakes are more likely so be careful while solving this problem.