Question 933275
Also 157+158.


More generally, suppose *[tex \large a + (a+1) + (a+2) + \ldots + (a+d) = 315] where *[tex \large d \ge 1]. The LHS equals *[tex \large a(d+1) + (1+2+\ldots + d) = a(d+1) + \frac{d(d+1)}{2} = 315] so *[tex \large (d+1)(a + \frac{d}{2}) = 315].


Multiply both sides by 2 to get *[tex \large (d+1)(2a+d) = 630]. Note that d+1 must necessarily be a factor of 630 that is at least 2. Once we fix d, we can uniquely determine a (note that d odd --> 2a+d odd, d even --> 2a+d even). The factors of 630 are: 1,2,3,5,6,7,9,10,14,15,18,21,30,35,42,45,63,70,90,105,126,210,315,630 (24 divisors), so d is 1 less than a factor of 630. The following list the values of a for each value of d:


d: a
0 315.0
1 157.0
2 104.0
4 61.0
5 50.0
6 42.0
8 31.0
9 27.0
13 16.0
14 14.0
17 9.0
20 5.0
29 -4.0
34 -8.0
41 -13.0
44 -15.0
62 -26.0
69 -30.0
89 -41.0
104 -49.0
125 -60.0
209 -103.0
314 -156.0
629 -314.0

For example, if d = 5, we have 50+51+...+(50+5) = 50+51+52+53+54+55+56 = 315. If d = 629, we have -314 + (-313) + ... + 315 = 315.

So in fact there are 24 possible ways to find consecutive numbers adding to 315 (including d = 0), but if we only include positive integers, and exclude d = 0, there are 11 ways total (d = 1, 2, ..., 17, 20).