Question 933275: I need help! I've tried several times to figure out this question and it just flusters me. "What are the sets of consecutive numbers that add up to 315?" I have found one and it is 104+105+106=315 and there are 11 total ways. I understand you are busy but i would really appriciate if you could help me out. Thank you.
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! "What are the sets of consecutive numbers that add up to 315?" I have found one and it is 104+105+106=315 and there are 11 total ways.
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It's the factors of 315
3 is one.
315/3 = 105, the middle # of the three --> 104, 105, 106
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Then 5.
315/5 = 63, the middle number --> 61, 62, 63, 64 & 65
then 7, 9, 15, 21, 35, 45, 63, 105 and 315.
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! Also 157+158.
More generally, suppose  + (a+2) + \ldots + (a+d) = 315) where  . The LHS equals  + (1+2+\ldots + d) = a(d+1) + \frac{d(d+1)}{2} = 315) so (a + \frac{d}{2}) = 315) .
Multiply both sides by 2 to get (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).
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