SOLUTION: In how many ways can the number 105 be written as the sum of two or more consecutive positive integers?

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Question 1204310: In how many ways can the number 105 be written as the sum of two or more consecutive positive integers?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
You want n+and r+such that
105=n%2B%28n%2B1%29%2B%28n%2B2%29+.....+%28n%2Br%29
=%28r%2B1%29n+(1%2B2+.....+r)
=%28r%2B1%29n%2Br%28r%2B1%29%2F2=%28%28r%2B1%29%282n%2Br%29%29%2F2,

which is equivalent to %28r%2B1%29%282n%2Br%29=210=2%2A3%2A5%2A7
.
Notice that, obviously, r%2B1%3C+2n%2Br, therefore %28r%2B1%29%5E2+%3C%28r%2B1%29%282n%2Br%29=105, whence r%2B1%3C=+abs%28sqrt%28210%29%29=14, so
r%2B1∈{ 2,3,5,2%2A3,7,2%2A5,2%2A7 }...(i.e. all the divisors of 105 that are ≤14- a total of highlight%287%29+cases).

here they are:

1. If r%2B1=2 then 2n%2Br=105; it follows that r=1 and n=52, so you have 105=52%2B53
.2 If r%2B1=3 then 2n%2Br=70; it follows that r=2 and n=34, so you have 105=34%2B35%2B36
.
3. If r%2B1=5 then 2n%2Br=42; it follows that r=4 and+n=19, so you have 105=19%2B20%2B21%2B22%2B23
.
4. If r%2B1=6 then 2n%2Br=35; it follows that+r=5 and n=15, so you have 105=15%2B16%2B17%2B18%2B19%2B20
.
5 If r%2B1=7+then 2n%2Br=30; it follows that r=6 and n=12, so you have
105=12%2B13%2B14%2B15%2B16%2B17%2B18
.
6. If r%2B1=10+then 2n%2Br=21; it follows that r=9 and+n=6, so you have 105=6%2B7%2B8%2B9%2B10%2B11%2B12%2B13%2B14%2B15
.
7. If r%2B1=14 then 2n%2Br=15; it follows that r=13 and n=1, so you have 105=1%2B2%2B3%2B4%2B5%2B6%2B7%2B8%2B9%2B10%2B11%2B12%2B13%2B14
.


Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.

The "solution" in the post by @MathLover 1 is one-to-one copy-paste from this web-page

https://math.stackexchange.com/questions/2088608/sum-as-an-increasing-sequence-of-two-or-more-consecutive-integers

without any reference, naturally.