# AMC 10 question

Question

In how many ways can $$345$$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

Factor $$345=3 \times 5 \times 23$$.

Suppose we take an odd number $$k$$ of consecutive integers, centered on $$m$$. Then $$mk=345$$ with $$\frac12k. Looking at the factors of $$345$$, the possible values of $$k$$ are $$3,5,15,23$$ centred on $$115,69,23,15$$ respectively.

Suppose instead we take an even number $$2k$$ of consecutive integers, centred on $$m$$ and $$m+1$$. Then $$k(2m+1)=345$$ with $$k\le m$$. Looking again at the factors of $$345$$, the possible values of $$k$$ are $$1,3,5$$ centered on $$(172,173),(57,58),(34,35)$$ respectively.

Thus the answer is $$\textbf{(E) }7$$.

Can someone explain to me how they got $$0.5 k < m$$ and the part saying centered on $$m$$ and $$m + 1$$.

• Well, with $k=2l+1$ the least element in your list is $m-l$ and we demand that this be positive.
– lulu
Commented May 9, 2017 at 11:25
• But what why add m and m+1?
– Math
Commented May 9, 2017 at 11:35
• I don't understand. An even numbered list has to have terms like $m,m+1$ in the middle. This is just a notation.
– lulu
Commented May 9, 2017 at 11:38

Observation: If you add an odd collection of $k$ consecutive integers, you get a multiple of the middle number. In other words, if you add $$(m-r)+(m-r+1)+\cdots+(m-2)+(m-1)+m+(m+1)+(m+2)+\cdots+(m+r),$$ there is a lot of cancellation, and you get $m(2r+1)$ as the sum, since there are $(2r+1)$ terms in the sum. To avoid nonpositive numbers, we need the smallest element in this list to be positive, so that $m-r>0$ or that $m>r$.

In your case, the total number of summands is $k=2r+1$ so $r=\frac{k-1}{2}$. Plugging this into the inequality above, you get: $$m>\frac{k-1}{2}.$$ The given inequality of $$m+\frac{1}{2}k$$ implies this necessary inequality.

Observation: If you add an even collection of $k$ integers, you get a multiple of the average of the middle two numbers. In other words, if you add $$(m-r)+\cdots+(m-2)+(m-1)+m+(m+1)+((m+1)+1)+((m+1)+2)+\cdots+((m+1)+r),$$ then you can pair up the terms to get $r+1$ copies of $m+(m+1)$. Therefore, this sum is $(2m+1)(r+1)$.

Once again, you need the smallest number in this list to be positive, so $m-r>0$ or that $m>r$. My $r$ differs from your $k$ by $1$, so the strict inequality becomes $\geq$ in your case.

Centered on $m$ and $m+1$ just means centered between those two numbers, i.e. $\frac{2m+1}{2}$. When $k=2$, your list is $\{m,m+1\}$, for $k=4$ your list is $\{m-1,m,m+1,m+2\}$, and so on.

For odd $k$, your list is centered on an integer $m$. When $k=3$, your list is $\{m-1,m,m+1\}$, for $k=5$ your list is $\{m-2,m-1,m,m+1,m+2\}$, and so on.

When summing k consecutive integers centered on $m$, your total $Y=km$. In your case, $Y=345$ and $m=3,5,15,$ or $23$ since those are the factors of 345.