Find the sum of all 3 digit number which leaves the remainder divided by 5. Show that for a odd integer
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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
hi
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
=99090
103, 108, 113,......, 998 is an A.P
First term of the A.P, a = 103
Common different of the A.P, d = 5
hi
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
=99090
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