Question

The sum of the integers from 1 to 100 that are divisible by 2 or 5 is
(a) 3000
(b) 3050
(c) 4050
(d) None of these

Answer 1

Here, we will write the series of the integers divisible by 2, divisible by 5, and divisible by 10. These series form arithmetic progressions. We will use the formula for ${n^{{\rm{th}}}}$ term of an A.P. and formula for sum of terms of an A.P. to find the sum of the three A.P. Finally, we will use these sums to find the sum of the integers from 1 to 100 that are divisible by 2 or 5.

Formula Used:
We will use the following formulas:
The ${n^{{\rm{th}}}}$ term of an A.P. is given by the formula ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the A.P. and $d$ is the common difference.
The sum of $n$ terms of an A.P. is given by the formula ${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$, where $a$ is the first term of the A.P. and $l$ is the last term of the A.P.

Complete step-by-step answer:
The integers divisible by 2 or 5 include the integers divisible by either 2, or 5, or both 2 and 5.
The integers from 1 to 100 that are divisible by 2 are 2, 4, 6, 8, 10, …, 96, 98, 100.
We can observe that the series 2, 4, 6, 8, 10, …, 96, 98, 100 is an arithmetic progression with first term 2 and common difference $4 - 2 = 2$.
We will find the sum of all the numbers from 1 to 100 that are divisible by 2 using the A.P.
First, we will find the number of terms of the A.P.
We can observe that the first term of the A.P. is 2, and the last term of the A.P. is 100.
Substituting ${a_n} = 100$, $a = 2$, and $d = 2$ in the formula for ${n^{{\rm{th}}}}$ term of an A.P., ${a_n} = a + \left( {n - 1} \right)d$, we get
$\Rightarrow 100 = 2 + \left( {n - 1} \right)2$
Subtracting 2 from both sides , we get
$\begin{array}{l} \Rightarrow 100 - 2 = 2 + \left( {n - 1} \right)2 - 2\\\ \Rightarrow 98 = \left( {n - 1} \right)2\end{array}$
Dividing both sides by 2, we get
$\begin{array}{l} \Rightarrow \dfrac{{98}}{2} = \dfrac{{\left( {n - 1} \right)2}}{2}\\\ \Rightarrow 49 = n - 1\end{array}$
Adding 1 to both sides, we get
$\begin{array}{l} \Rightarrow 49 + 1 = n - 1 + 1\\\ \Rightarrow 50 = n\end{array}$
$\therefore$ The number of terms in the A.P. 2, 4, 6, 8, 10, …, 96, 98, 100 is 1001.
Now, we will find the sum of all terms in the A.P. 2, 4, 6, 8, 10, …, 96, 98, 100.
Let ${S_1}$ be the sum of all terms in the A.P. 2, 4, 6, 8, 10, …, 96, 98, 100.
Substituting $n = 50$, $a = 2$, and $l = 100$ in the formula for sum of $n$ terms of an A.P., ${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$, we get
$\Rightarrow {S_1} = \dfrac{{50}}{2}\left[ {2 + 100} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {S_1} = \dfrac{{50}}{2}\left[ {102} \right]\\\ \Rightarrow {S_1} = 25 \times 102\\\ \Rightarrow {S_1} = 2550\end{array}$
Now, the integers from 1 to 100 that are divisible by 5 are 5, 10, 15, …, 95, 100.
We can observe that the series 5, 10, 15, …, 95, 100 is an arithmetic progression with first term 5 and common difference $10 - 5 = 5$.
We will find the sum of all the numbers from 1 to 100 that are divisible by 5 using the A.P.
First, we will find the number of terms of the A.P.
We can observe that the first term of the A.P. is 5, and the last term of the A.P. is 100.
Substituting ${a_n} = 100$, $a = 5$, and $d = 5$ in the formula for ${n^{{\rm{th}}}}$ term of an A.P., we get
$\Rightarrow 100 = 5 + \left( {n - 1} \right)5$
Subtracting 5 from both sides of the equation, we get
$\begin{array}{l} \Rightarrow 100 - 5 = 5 + \left( {n - 1} \right)5 - 5\\\ \Rightarrow 95 = \left( {n - 1} \right)5\end{array}$
Dividing both sides of the equation by 5, we get
$\begin{array}{l} \Rightarrow \dfrac{{95}}{5} = \dfrac{{\left( {n - 1} \right)5}}{5}\\\ \Rightarrow 19 = n - 1\end{array}$
Adding 1 to both sides of the equation, we get
$\begin{array}{l} \Rightarrow 19 + 1 = n - 1 + 1\\\ \Rightarrow 20 = n\end{array}$
Therefore, the number of terms in the A.P. 5, 10, 15, …, 95, 100 is 20.
Now, we will find the sum of all terms in the A.P. 5, 10, 15, …, 95, 100.
Let ${S_2}$ be the sum of all terms in the A.P. 5 are 5, 10, 15, …, 95, 100.
Substituting $n = 20$, $a = 5$, and $l = 100$ in the formula for sum of $n$ terms of an A.P., ${S_n} = \dfrac{n}{2}\left[ {a + l} \right]$, we get
$\Rightarrow {S_2} = \dfrac{{20}}{2}\left[ {5 + 100} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {S_2} = 10\left[ {105} \right]\\\ \Rightarrow {S_2} = 1050\end{array}$
Now, we will find the integers divisible by both 2 and 5.
The integers divisible by both 2 and 5 are the numbers divisible by 10.
The integers from 1 to 100 that are divisible by 10 are 10, 20, …, 90, 100.
We can observe that the series 10, 20, …, 90, 100 is an arithmetic progression with first term 10 and common difference $20 - 10 = 10$.
We will find the sum of all the numbers from 1 to 100 that are divisible by 10 using the A.P.
First, we will find the number of terms of the A.P.
We can observe that the first term of the A.P. is 10, and the last term of the A.P. is 100.
Substituting ${a_n} = 100$, $a = 10$, and $d = 10$ in the formula for ${n^{{\rm{th}}}}$ term of an A.P., we get
$\Rightarrow 100 = 10 + \left( {n - 1} \right)10$
Subtracting 10 from both sides of the equation, we get
$\begin{array}{l} \Rightarrow 100 - 10 = 10 + \left( {n - 1} \right)10 - 10\\\ \Rightarrow 90 = \left( {n - 1} \right)10\end{array}$
Dividing both sides of the equation by 10, we get
$\begin{array}{l} \Rightarrow \dfrac{{90}}{{10}} = \dfrac{{\left( {n - 1} \right)10}}{{10}}\\\ \Rightarrow 9 = n - 1\end{array}$
Adding 1 to both sides of the equation, we get
$\begin{array}{l} \Rightarrow 9 + 1 = n - 1 + 1\\\ \Rightarrow 10 = n\end{array}$
Therefore, the number of terms in the A.P. 10, 20, …, 90, 100 is 10.
Now, we will find the sum of all terms in the A.P. 10, 20, …, 90, 100.
Let ${S_3}$ be the sum of all terms in the A.P. 10, 20, …, 90, 100.
Substituting $n = 10$, $a = 10$, and $l = 100$ in the formula for sum of $n$ terms of an A.P., we get
$\Rightarrow {S_3} = \dfrac{{10}}{2}\left[ {10 + 100} \right]$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {S_3} = 5\left[ {110} \right]\\\ \Rightarrow {S_3} = 550\end{array}$
Now, we will find the sum of all integers from 1 to 100 that are divisible by 2 or 5.
We can observe that the sum of the integers that are divisible by 10 are already included in the sum of the integers divisible by 2, and the sum of integers divisible by 5.
Thus, if we add ${S_1}$ and ${S_2}$, the sum of integers divisible by 10 would be included twice.
We will subtract the sum of integers divisible by 10 once to get the correct answer.
Therefore, we get
Sum of all integers from 1 to 100 that are divisible by 2 or 5 $= {S_1} + {S_2} - {S_3}$
Substituting ${S_1} = 2550$, ${S_2} = 1050$, and ${S_3} = 550$ in the expression, we get
Sum of all integers from 1 to 100 that are divisible by 2 or 5 $= 2550 + 1050 - 550$
Simplifying the expression, we get
Sum of all integers from 1 to 100 that are divisible by 2 or 5 $= 3050$
Therefore, we get the sum of all integers from 1 to 100 that are divisible by 2 or 5 as 3050.
Thus, the correct option is option (b).

Note: Here, we need to understand why the sum of integers divisible by 2 and 5 is deducted. A common mistake is to add the sum of integers divisible by 2 or 5, that is 550. This results in the answer 4050, which is incorrect. We can also find the sum using the common difference of the A.P. using the formula ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $a$ is the first term of the A.P., $d$ is the common difference of the A.P. and $n$ is the number of terms of the A.P.