50 exclaimed with joy

Number Theory Level 5

In how many ways can $50!$ be expressed as a sum of two or more consecutive positive integers?

The answer is 93000959.

1 solution

Otto Bretscher
Aug 27, 2015

According the theory for these matters, we need to count the number of odd factors of $50!$ . Each such factor gives rise to a representation of $50!$ as a sum of consecutive positive integers.

Now $50!=2^{47}\times3^{22}\times5^{12}\times7^8\times11^4\times13^3\times17^2\times19^2\times23^2$ $\times29\times31\times37\times41\times43\times47$ . Thus the number of odd factors is $23*13*9*5*4*3^3*2^6=93000960$ . The number of representations we seek is one less, $\boxed{93000959}$ , since we exclude the trivial representation $50!=50!$ .

Can you provide a link to "the theory for these matters"? I'm interested in reading about this some more.

Ryan Tamburrino - 5 years, 9 months ago

Here is a pretty lucid discussion, carefully explaining the parity issues: http://2000clicks.com/mathhelp/PuzzleSumsOfConsecutiveIntegersAnswer.aspx

Otto Bretscher - 5 years, 9 months ago

Let me explain this in my own words: If $m$ is an odd factor of $n$ , then we can write $n$ as the sum of the $m$ consecutive integers with middle term $n/m$ . If the first term of this sum is negative, $-p$ , we can cancel the terms $-p+...+0+...+p$ in the sum to write $n$ as a sum of consecutive positive integers. Conversely, it is not hard to see that all representations of $n$ as sums of consecutive positive integers are obtained in this way.

Let's illustrate this with the example $n=15$ , with the odd factors $1,3,5,$ and $15$ . We have the representations $15=15$ $=4+5+6=1+2+3+4+5$ $=-6-5+...+0+1+...+6+7+8=7+8.$

Otto Bretscher - 5 years, 9 months ago

50 exclaimed with joy

The answer is 93000959.

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