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`        Find the sum of all three digit numbers which leave remainder 2 when divided by 5.  (A) 98910 (B) 9820(C) 9830 (D) 9840`
4 months ago

```							Three digit numbers which leave the remainder 2 when divided by 5 are 102, 107, 112,117......, 997.102, 107, 112,......, 997 is an A.PIn this A.P a (first term)= 102d (Common difference)= 5I(last term ) = 997 l= an = a + (n – 1) d997= 102 + (n – 1) × 55 (n – 1) = 997 – 102 = 895n-1= 885/5(n – 1) = 179n = 179 +1n = 180Sum of all three digit numbers which leaves remainder 2 when divided by 5Sn = n/2 [a+l]Sn= 180/2 [ 102+ 997]Sn= 90 × 1099Sn= 98910Sum of all three digit numbers which leaves remainder 2 when divided by 5 is 98910
```
4 months ago
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