Question

How many three digit natural number will leave remainder 3 when divisible 5.​

Answer 1

Step-by-step explanation:

$\red{Solution》}$

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

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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

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

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NOTE:- THE SYMBOL "__" SHOW THE CONTINUITY