Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

  • -12

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (n –1) 2

n = 50

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

∴100 = 5 + (n –1) 5

⇒ 5n = 100

n = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

∴100 = 10 + (n –1) (10)

⇒ 100 = 10n

n = 10

∴Required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

  • 110

Find the sum of numbers from 1 to 100 which are neither divisible by 2 nor by 5.

THEN ANSWER WILL BE -

The sum of all numbers which are divisible by 2 or 5 = sum of all numbers divisible by 2 + sum of all numbers divisible by 5 - sum of all numbers divisible by both 2 and 5 

Firstly we will find the sum of all numbers between 1 to 100 which are divisible by 2 as: 

The numbers divisible by 2 are: 2, 4, 6, 8, ......., 100.

Therefore, an = a + (n - 1)d

⇒ 100 = 2 + (n - 1)2

⇒ 98/2 = (n - 1)

⇒ n = 49 + 1 = 50

Similarly, 

The numbers divisible by 5 are: 5, 10, 15,  ......., 100.

Therefore, an = a + (n - 1)d

⇒ 100 = 5 + (n - 1)5

⇒ 95/5 = (n - 1)

⇒ n = 19 + 1 = 20

Again, 

The numbers divisible by both 2 and 5 or multiples of 10 are : 10, 20,  ......., 100.

Therefore, an = a + (n - 1)d

⇒ 100 = 10 + (n - 1)10

⇒ 90/10 = (n - 1)

⇒ n = 9 + 1 = 10

Sum of all numbers divisible by 2 or 5 = S50 + S20 - S10 = 2550+1050 -1100 = 2500

Again, sum of all numbers from 1 to 100

Now, sum of all numbers neither divisible by 2 nor by 5 =  sum of all numbers from 1 to 100 - Sum of all numbers divisible by 2 or 5

= 5050 - 2500 = 2550

  • 7
What are you looking for?