Question

Find the sum of all three-digit natural numbers which are divisible by 13.

Answer 1

Smallest three digit number = 100
Largest three digit number = 999

104 is divisible by 13 , and the largest 988 is divisible by 13 .

Sum of all three digit number divisible by 13

= 104 + 117 + 130 + 143 + 156 + 169 + ... + 988

= 13 ( 8 + 9 + 10 + 11 + 12 + 13 + ... + 76 )

= 13 [ ( 1 + 2 + 3 + ... + 76 ) - ( 1 + 2 + 3 + ... + 7 ) ]

= 13 [ ( 76)(77)/2 - 7(8)/2 ]

= 13 [ 2926 - 28 ]

= 13 ( 2898 )

= 37,674

Therefore, The sum of all three digit natural numbers which are divisible by 13 is 37674 .

I used the formula of sum of first n natural numbers in this process.

Answer 2

⇒ First three digit number divisible by 13 = 104.

⇒ Last three digit number divisible by 13 = 988.

So, the series starts from 104 and ends with 988.

The series will be:

104, 117,130,143...988

Since difference is Same, it is an AP.

First term a = 104.

Common difference d = 13.

Last term an = l = 988.

Now,

We need to find number of terms, i.e n

⇒ an = a + (n - 1) * d

⇒ 988 = 104 + (n - 1) * 13

⇒ 988 = 104 + 13n - 13

⇒ 988 = 13n + 91

⇒ 988 - 91 = 13n

⇒ 897 = 31n

⇒ n = 69.

We know that sum of first n terms in an AP is:

⇒ sn = n/2[a + l]

        = 69/2[104 + 988]

        = 69/2[1092]

        = 69 * 546

        = 37674.

Therefore, Sum of all three digit natural numbers divisible by 13 is 37674.

Hope this helps!