By the fundamental theorem of arithmetic, every positive integer greater than 1 can be written as a product of primes. Thus, we can write every positive integer as $2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$ for some non-2 primes $\{ p_{ i }\} _{ i=1 }^{ m }$, some non-negative integer $k$ and some positive integers $\{ q_{ i }\} _{ i=1 }^{ m }$. Let's say that the sequence of consecutive integers starts at some integer $x$ and ends at some integer $x+n$ where $n$ is some positive integer (thus there are $n+1$ terms in the sequence). Note that this excludes sets with just one member. The sum of the consecutive integers is then given by $\sum _{ k=0 }^{ n }{ (x+k) }$, and thus we need $\sum _{ k=0 }^{ n }{ (x+k) } =2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$ for the particular $x$ and $n$ to constitute a valid sequence of $n+1$ members. This gives that $(n+1)x+\frac { n(n+1) }{ 2 } =2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad \Leftrightarrow \quad x+\frac { n }{ 2 } =\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$, and thus $x=\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 } -\frac { n }{ 2 }$.

So, the question becomes, how many positive integers $n$ can we choose such that the right-hand side of that equation is an integer and thus, so too, is $x$.

Well, there are two potential cases, that $n$ is even and that $n$ is odd. If $n$ is even, we have that $\frac { n }{ 2 }$ is an integer and thus, so too, must $\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$ be. Therefore, we have that $(n+1)|2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$. If that is the case, along with n being even, $x$ must then be an integer and $x$ and $n$ constitute a valid pair. Thus, $n+1$ must be any divisor of $2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$ such that $n$ is even, or, equivalently that $n+1$ is odd. Thus, we can let $n+1$ be any odd, positive divisor (except 1) of $2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$ giving $(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)-1$ valid pairs $(x,n)$ which have even $n$ (we wish to exclude the case where $n+1=1$, as $n$ must be a positive integer). Next, we have the case that $n$ is odd. Going back to $x+\frac { n }{ 2 } =\frac { 2^{ k }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }\quad }{ n+1 }$, we have, multiplying both sides by 2, that $2x+n=\frac { 2^{ k+1 }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } } }{ n+1 }$. Noting that because $n$ is odd and $2x$ is even, the left-hand side is an odd integer, and thus, so too, must the right-hand side be. Thus, $n+1$ must be a divisor of $\frac { 2^{ k+1 }p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } } }{ n+1 }$ such that $2^{ k+1 }|(n+1)$. Therefore, we have that $n+1=2^{ k+1 }d$ where $d$ can be any positive divisor of $p_{ 1 }^{ q_{ 1 } }p_{ 2 }^{ q_{ 2 } }...p_{ m }^{ q_{ m } }$, as if $d$ is a positive integer and $k$ is a non-negative integer $2^{ k+1 }d-1$ will always be a positive integer. Thus, because each $d$ corresponds to $1$ valid odd $n$ we have $(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)$ valid odd $n$ and, thus, $2(q_{ 1 }+1)(q_{ 2 }+1)(q_{ 3 }+1)...(q_{ m }+1)-1$ total valid $n$, corresponding to exactly that many sequences of consecutive integers which sum to our original positive integer (excluding the sequence of just that positive integer)!

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TopNewestThis is an unusually interesting setup. I think having a problem that applies these stuffs might be worthwhile. Would you like to post one?

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