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# Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5.

Last updated date: 23rd Jul 2024
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Hint- Here, the formulas of ${n^{{\text{th}}}}$ term of an AP and sum of first $n$ terms of an AP will be used.
The natural numbers which are divisible by 2 are 2, 4, 6, 8, ……, 98, 100.
Clearly, the above series is an arithmetic progression with a common difference of 2.
The natural numbers which are divisible by 5 are 5, 10, 15, 20, ……, 95, 100.
Clearly, the above series is an arithmetic progression with a common difference of 5.
For an arithmetic progression consisting of first term as $a$, common difference as $d$ and total number of terms as $n$
Last term or ${n^{{\text{th}}}}$ term of the AP is given by ${a_n} = a + \left( {n - 1} \right)d{\text{ }} \to {\text{(1)}}$
Sum of first $n$ terms of the AP is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {a + {a_n}} \right]{\text{ }} \to {\text{(2)}}$.
Now, for the arithmetic progression 2, 4, 6, 8, ……, 98, 100
$a = 2$, $d = 2$ and ${a_n} = 100$
Using equation (1), we can write
$\Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 2 + 2n - 2 \Rightarrow 2n = 100 \Rightarrow n = 50$
So, the total number of natural numbers between 1 and 100 that are divisible by 2 are 50.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{50}}{2}\left[ {2 + 100} \right] = \dfrac{{50 \times 102}}{2} = 2550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 2 is 2550.
Now, for the arithmetic progression 5, 10, 15, 20, ……, 95, 100
$a = 5$, $d = 5$ and ${a_n} = 100$
Using equation (1), we can write
$\Rightarrow 100 = 2 + 2\left( {n - 1} \right) \Rightarrow 100 = 5 + 5\left( {n - 1} \right) \Rightarrow 100 = 5 + 5n - 5 \Rightarrow 5n = 100 \Rightarrow n = 20$
So, the total number of natural numbers between 1 and 100 that are divisible by 5 are 20.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{20}}{2}\left[ {5 + 100} \right] = \dfrac{{20 \times 105}}{2} = 1050$
So, the sum of all the natural numbers between 1 and 100 that are divisible by 5 is 1050.
Also, natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10, 20, 30, ……, 100
The above series also forms arithmetic progression with a common difference of 10.
Here, $a = 10$, $d = 10$ and ${a_n} = 100$
Using equation (1), we can write
$\Rightarrow 100 = 10 + 10\left( {n - 1} \right) \Rightarrow 90 = 10n - 10 \Rightarrow 10n = 100 \Rightarrow n = 10$
So, the total number of natural numbers between 1 and 100 that are divisible by both 2 and 5 are 10.
Now using equation (2), we have
${{\text{S}}_n} = \dfrac{{10}}{2}\left[ {10 + 100} \right] = \dfrac{{10 \times 110}}{2} = 550$
So, the sum of all the natural numbers between 1 and 100 that are divisible by both 2 and 5 is 550.
Since, the sum of all the natural numbers between 1 and 100 that are divisible by 2 or 5 is equal to the sum of all the natural numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 minus the number of natural numbers between 1 and 100 that are divisible by both 2 and 5.
i.e., ${\text{Required Sum}} = \left( {2550 + 1050} \right) - 550 = 3050$.
Therefore, the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5 is 3050.

Note- In this particular problem, the sum of numbers between 1 and 100 that are divisible by 2 and those that are divisible by 5 will include the numbers that are divisible by both 2 and 5 twice. So in order to find the accurate result we will once subtract the sum of all the numbers between 1 and 100 that are divisible by both 2 and 5.