Question

Find the sum of all three digit numbers which leave 3 On dividing by 5​

Answer 1

the answer is

99090

hope you like this answer

Answer 2

Answer:

Step-by-step explanation:

Three digit numbers which leave the remainder 3 when divided by 5 are 103, 108, 113,......., 998.

103, 108, 113,......, 998 is an A.P

First term of the A.P, a = 103

Common different of the A.P, d = 5

Let 998 be the nth term of the A.P.

an = a + (n – 1) d

∴ 103 + (n – 1) × 5 = 998

⇒ 5 (n – 1) = 998 – 103 = 895  

⇒ (n – 1) = 179

⇒  n = 180

Sum of all three digit numbers which leaves remainder 3 when divided by 5

180/2 ( 103 + 998 ) = 90 ( 1101 ) = 99090