Question

find the sum of all two digit natural numbers which is divisible by 13​

Answer 1

Hence, sum if 2- digit number which when divided by 3 yied 1 as remainder is 1605

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Answer 2

Answer:

The smallest and the largest of three digits which are divisible by 13 are 104 and 988 respectively.

So, the sequence of three digits numbers which are divisible by 13 are ------>

104 , 117, 130,......988

It is an A.P with first term, a = 4 and common difference, d =13.

let, there be n terms in this sequence, then

$a_{n} = 988 \\\implies \: a + (n - 1)d = 988 \\ \implies104 + (n - 1)13 = 988 \\ \implies104 + 13n \: - 13 = 988 \\ \implies \: n = 69$

$∴ required \: sum ,  s_{n} = \frac{n}{2} [2a + (n - 1)d] \\ = \frac{69}{2} [2 \times 104 + (69 - 1)13] \\ = \frac{69}{2} \times 1092 \\ = 69 \times 546 \\ = 37674$