Question
The sum of all the 3-digit numbers, each of which on division by 5 leaves remainder 3, is
Answer: Option D
Answer: (d)According to the question, First number = a = 103; Last number = l = 998 If the number of such numbers be n, then, 998 = 103 + (n – 1) × 5 (n – 1) × 5= 998 – 103 = 895 n – 1 = $895/5$ = 179n = 180S = $n/2(a+l)$=$180/2$(103 +998) = 90 × 1101 = 99090
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Answer: (d)According to the question, First number = a = 103; Last number = l = 998 If the number of such numbers be n, then, 998 = 103 + (n – 1) × 5 (n – 1) × 5= 998 – 103 = 895 n – 1 = $895/5$ = 179n = 180S = $n/2(a+l)$=$180/2$(103 +998) = 90 × 1101 = 99090
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