Question
Download Solution PDFFind the sum of all 3 digit numbers that leave a remainder of 5 when divided by 7.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
We need to find the sum of all 3-digit numbers which leave a remainder of 5 when divided by 7.
Formula used:
Sum of AP: S = n/2 × (first term + last term)
Calculations:
General form: number = 7k + 5
Smallest 3-digit number = 100, Largest = 999
First such number: 7 × 14 + 5 = 103
Last such number: 7 × 142 + 5 = 999
This is an Arithmetic Progression (AP): a = 103, l = 999, d = 7
Number of terms: n = ((l - a)/d) + 1 = ((999 - 103)/7) + 1 = 896/7 + 1 = 128 + 1 = 129
Sum of AP: S = n/2 × (first term + last term)
S = 129/2 × (103 + 999) = 129/2 × 1102 = (129 × 1102) ÷ 2 = 142158 ÷ 2 = 71079
∴ Sum of all such 3-digit numbers = 71,079
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