Find the sum of all the numbers in the range: [100, 999], which leave a remainder of 3 when divided by prime number 7.
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70821
60821
50521
80821
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50521 . of the season finale was defeated by a car accident
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The sum of all such numbers would be 70821
1) The first number from 100 which when divided by 7 leaves a remainder of 3 is 101 i.e. 14*7 +3.
2) The subsequent numbers from 101 would be 108, 115, 122, 129,......997
3) 997 can be written as 142*7 +3
4) Therefore, total numbers would be 142-14+1 = 129
5) The sum of a GP is given by
Sₙ = n/2(a₁ + aₙ)
6) Here, a₁ is 101, aₙ is 997 and n is 129
7) Sum of AP would be 129/2(101+997) = 129* 1098/2
= 70821
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