Elite Consecutive Integers

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 = 93000959 ways to express 50! as the sum of two or more consecutive positive integers. 9 + 3 + 0 + 0 + 0 + 9 + 5 + 9 = 35

Let $k$ denote the length of the string of consecutive integers that are to sum to $50!$

Consider the case where $k$ is even.

For any string of $k$ consecutive integers, one must be divisible by $k$ , and so the string is equivalent to a permutation of $1,2,...,k-1,0 \pmod{k}$ . The sum is therefore equivalent to $\sum_{a=1}^{k-1} a = \frac{1}{2}\left(k-1\right)\left(k\right) \pmod{k}$

This implies that $\frac{50!}{k} \equiv \frac{k-1}{2} \pmod{k}$ Since $k-1$ is odd, this must mean that $k$ is $2b*2^a$ where $a$ is the largest power of $2$ that divides $50!$ , and b is some odd number that divides $50!$

This therefore means that for every $b$ that divides $50!$ there exists a string of $k$ consecutive integers that add up to $50!$

It now remains to find out how many odd factors of $50!$ there are which can be done by counting the powers of various primes in the numbers $1$ to $50$ . Doing so gives us that the largest odd factor of $50!$ is:

$3^{22}*5^{12}*7^8*11^4*13^3*17^2*19^2*23^2*29^1*31^1*37^1*41^1*43^1*47^1$

Therefore, by the divisor function we see that there are $23*13*9*5*4*3*3*3*2*2*2*2*2*2 -1 = 93000960$ odd factors of $50!$ .

Now for some odd $k$ we must have a string of $k$ consecutive integers that add up to $50!$ As each number, roughly speaking, contributes $\frac{1}{k}th$ of the sum, the numbers must therefore be clustered around the number $\frac{50!}{k}$

An odd number of consecutive integers can be written as $\left(a-b\right), \left(a-b+1\right) ..., a, ...., \left(a+b-1\right), \left(a+b\right)$ , which sums up to $\left(2b+1\right)a$

This implies that $\frac{50!}{k}$ must be the "centre value" or else the symmetry doesn't apply and the "pluses" and "minuses" don't completely cancel out. Therefore, for every $k: k = 2n-1, n\in \mathbb{Z^+}, k|50!$ , there exists a string of $k$ consecutive positive integers that add up to $50!$

So now we have $2*93000960$ strings of consecutive integers that add up to $50!$ , however some strings contain negative numbers when the central value or median $a$ is such that $a < \frac{k+1}{2}$ However, due to symmetry around $0$ and the fact that no two strings have the same length and thus contain a different set of integers, it must be true that every string with negative numbers can be mapped onto a unique string of positive integers that add up to $50!$ , and hence there must be $93000960$ strings of numbers that add up to $50!$

One of these strings is of length $k=1$ , i.e, $50!$ , so there are $93000960-1 = 93000959$ strings of length $k \geq 2$

Hence, the answer we want is $9+3+9+5+9 = 35$

The number of odd number factor of $50!$ is $93000959$

$\fbox{35}$

For any given number , say 100 the number of ways 100 can be expressed as a sum of consecutive positive integers is : [No.of odd factors - 1]. The logic is as follows : 100 = n/2 (2a+(n-1)d); d= 1. n (2a+n-1) = 200 = even odd product . Factorizing 200 into odd even products gives the different values of n and a and hence, the no. of ways. For fact(50) : find the exponents of 3,5,7,11 and so on and find the no. of factors. Ans -1 gives the value of w = 93000959 Sum of digits = 35