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In how many ways can $105$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

I got the above question for small numbers. For example I took 15.

Since $15=3\times5$

$15=3+3+3+3+3=(3-2)+(3-1)+(3)+(3+1)+(3+2)=1+2+3+4+5$

$15=5+5+5=(5-1)+(5)+(5+1)=4+5+6$

So there are $2$ ways for $15.$ Can I apply this method to $105.$ How ?

  • 1
    You missed 7+8, – 2017-01-08
  • 0
    See whether [this helps you](http://math.stackexchange.com/questions/139842/in-how-many-ways-can-a-number-be-expressed-as-a-sum-of-consecutive-numbers). Also see [here](https://mathblag.wordpress.com/2011/11/13/sums-of-consecutive-integers/). – 2017-01-08

1 Answers 1

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You want $n$ and $r$ such that

$$105 = n + (n+1) + (n+2) + \dots + (n+r) = (r+1)n + (1 + 2 + \dots + r) = \\(r+1)n + \frac {r(r+1)} 2 = \frac {(r+1) (2n+r)} 2 ,$$

which is equivalent to $(r+1) (2n+r) = 210 = 2 \cdot 3 \cdot 5 \cdot 7$.

Notice that, obviously, $r+1 < 2n+r$, therefore $(r+1)^2 < (r+1)(2n+r) = 105$, whence $r+1 \le [\sqrt {210}] = 14$, so $r+1 \in \{2, 3, 5, 2 \cdot 3, 7, 2 \cdot 5, 2 \cdot 7 \}$ (i.e. all the divisors of $105$ that are $\le 14$ - a total of 7 cases).

  1. If $r+1 = 2$ then $2n + r = 105$; it follows that $r=1$ and $n=52$, so you have $105 = 52 + 53$.

  2. If $r+1 = 3$ then $2n + r = 70$; it follows that $r=2$ and $n=34$, so you have $105 = 34 + 35 + 36$.

  3. If $r+1 = 5$ then $2n + r = 42$; it follows that $r=4$ and $n=19$, so you have $105 = 19 + 20 + 21 + 22 + 23$.

  4. If $r+1 = 6$ then $2n + r = 35$; it follows that $r=5$ and $n=15$, so you have $105 = 15 + 16 + 17 + 18 + 19 + 20$.

  5. If $r+1 = 7$ then $2n + r = 30$; it follows that $r=6$ and $n=12$, so you have $105 = 12 + 13 + 14 + 15 + 16 + 17 + 18$.

  6. If $r+1 = 10$ then $2n + r = 21$; it follows that $r=9$ and $n=6$, so you have $105 = 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15$.

  7. If $r+1 = 14$ then $2n + r = 15$; it follows that $r=13$ and $n=1$, so you have $105 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14$.

  • 0
    Also setting $r+1=2\cdot3$ and $r+1=2\cdot5$ gives you $n=15$ and $n=6$ respectively. – 2017-01-08
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    What if r+1=6, r=5,n=15. That is also another case. So are there 5 cases. Thanks for the answer. – 2017-01-08
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    @AlexM. See this one is this useless http://math.stackexchange.com/questions/414022/how-to-find-a-nonzero-2-times-2-matrix-whose-square-is-zero/2087508#2087508 – 2017-01-08
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    @AlexM. Also see this one http://math.stackexchange.com/questions/2087364/limit-of-the-greatest-integer-of-frac100-tanx-sinxx2-as-x-tends – 2017-01-08
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    @AlexM. Also see this one http://math.stackexchange.com/questions/414022/how-to-find-a-nonzero-2-times-2-matrix-whose-square-is-zero/2061335#2061335 – 2017-01-08
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    @AlexM. But I didn't give the answer but I just improved formatting of the first post of a new user and I have the privilege to delete posts – 2017-01-08
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    @AlexM. also see this one http://math.stackexchange.com/questions/2088566/can-a-periodic-function-be-represented-with-roots-x2-where-x-is-an-elemen – 2017-01-08
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    @AlexM. also see this http://math.stackexchange.com/questions/2087114/find-bd-bc-dc – 2017-01-08