Consecutive sum

We need to find all positive $n,k$ such that $n + (n + 1) + .... + (n + k) = 100$ . We can rewrite this equation as

$(1 + 2 + .... + (n - 1) + n + .... + (n + k)) - (1 + 2 + ... + (n - 1)) = 100 \Longrightarrow \dfrac{(n + k)(n + k + 1)}{2} - \dfrac{(n - 1) \times n}{2} = 100 \Longrightarrow$

$(n + k)^{2} + (n + k) - n(n - 1) = 200 \Longrightarrow n^{2} + k^{2} + 2nk + n + k - n^{2} + n = 200 \Longrightarrow k^{2} + k + 2nk + n = 200 \Longrightarrow$

$k(k + 1) + 2n(k + 1) = 200 \Longrightarrow (k + 2n)(k + 1) = 200$ .

Now consideration the possible two-element factorizations of $200$ , noting that as $n \ge 1$ we must have $k + 2n \gt k + 1$ . These factorizations are

• (i) $k + 2n = 200, k + 1 = 1 \Longrightarrow k = 0, n = 100$ , which would only involve one term so we can discard this option,

• (ii) $k + 2n = 100, k + 1 = 2 \Longrightarrow k = 1, n = 49.5$ , but we require a sequence of integers, so discard,

• (iii) $k + 2n = 50, k + 1 = 4 \Longrightarrow k = 3, n = 23.5$ , discard as for (ii),

• (iv) $k + 2n = 40, k + 1 = 5 \Longrightarrow k = 4, n = 18$ , giving us the valid sequence $18, 19, 20, 21, 22$ ,

• (v) $k + 2n = 25, k + 1 = 8 \Longrightarrow k = 7, n = 9$ , giving us the valid sequence $9, 10, 11, 12, 13, 14, 15, 16$ ,

• (vi) $k + 2n = 20, k + 1 = 10 \Longrightarrow k = 9, n = 5.5$ , which would not involve a sequence of integers.

So the two valid sequences begin with $18$ and $9$ respectively, giving us a desired answer of $18 + 9 = \boxed{27}$ .

Let the required set begins with $a$ and ends in $b$ , where $a$ and $b$ are positive integers. Then the sum of the elements of the set is $\dfrac{b(b+1)}{2}-\dfrac{a(a+1)}{2}$ . Therefore $b^2-a^2+b-a=200$ , or $(b-a) (b+a+1)=200$ . Now $200$ is the product of (i) $1$ and $200$ , (ii) $5$ and $40$ , and (iii) $25$ and $8$ . ( $b-a$ and $b+a+1$ are of opposite parity). Hence (i) $a=99$ , $b=100$ , (ii) $a=17$ , $b=22$ , and (iii) $a=8$ , $b=16$ . Therefore the sets are { $100$ }, { $18,19,20,21,22$ } and { $9,10,11,12,13,14,15,16$ }. Hence the required answer is $100+18+9=\boxed {127}$