Question

What is the sum of integers from \(113\) to \(113113\) that are divisible by \(7\)?

• A. 16143
• B. 113113
• C. 16159
• D. 913952088

Hint:

For “sum of multiples in a range,” pick the first and last multiples, count terms by \(\frac{l-a}{d}+1\), then use \(S=\frac{n}{2}(a+l)\).

Answer 1

The Correct Option is D

First multiple of \(7\) \(\ge 113\) is \(119\). Last multiple \(\le 113113\) is \(113113\) itself (since \(113113=7\times 16159\)).
This is an AP with first term \(a=119\), last term \(l=113113\), common difference \(d=7\).
Number of terms \[ n=\frac{l-a}{d}+1=\frac{113113-119}{7}+1=16143. \] Sum \[ S=\frac{n(a+l)}{2}=\frac{16143\,(119+113113)}{2} =\frac{16143\times 113232}{2}=913952088. \] \[ \boxed{913952088} \]