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

The answer is 93000959.

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A number can be expressed as the sum of an odd number of consecutive integers if the middle number is an integer. There are 93000960 ways to do that (number of odd factors of 50!). For each odd factor of 50!, you can divide 50! by that factor and the resulting number is the middle number and the odd factor is the number of terms. A number can be expressed as the sum of an even number of consecutive integers if the mean of the numbers is an integer with a .5 after it. That means that if you divide 50! by the number of terms, you end up with a number that ends with .5, so the prime factorization of all of the possible number of terms has $2^{48}$ in it (50! has $2^{47}$ in its prime factorization, so dividing by $2^{48}$ will give the .5). The number of ways to get an even number of consecutive integers to add to 50! is the amount of numbers that could divide 50! and result in a number with a .5. This is equivalent to the number of odd factors of 50!, and each odd factor multiplied by $2^{48}$ are the numbers we're looking for. Dividing 50! by the odd factor would still let 50! keep the $2^{47}$ part of its prime factorization and let it remain an integer as well, and the $2^{48}$ will let the quotient have .5 in the end. There are $93000960 * 2$ ways to express 50! as the sum of consecutive integers, but only half of those have all positive integers. Each way of positive integers can be matched up with a pair with negative integers, since all possible ways were counted. Ex: -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 = 3 + 4 + 5 (the -2 to 2 part of the left side cancels out). We are now left with 93000960 ways to represent 50! as the sum of consecutive integers, and one of those ways has one positive integer (50! by itself), so there are 93000960 - 1 = $\boxed{93000959}$ ways to express 50! as the sum of two or more consecutive positive integers.