## It is possible to write five as a sum in exactly six different ways:

\begin{align}

&4 + 1\\

&3 + 2\\

&3 + 1 + 1\\

&2 + 2 + 1\\

&2 + 1 + 1 + 1\\

&1 + 1 + 1 + 1 + 1

\end{align}

How many different ways can one hundred be written as a sum of at least two positive integers?

### This problem is a classic type of problem in number theory, particularly partition theory, that often asks for the number of partitions of an integer.

A partition of an integer $n$ is a way of writing $n$ as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition.

The number of partitions of 100 is a well-studied problem, but to compute it exactly, one typically uses a slightly sophisticated technique involving generating functions. However, calculating it by hand could be very time-consuming and not practical.

Thankfully, part of number theory and number partition theory has built on these kind of solutions, and the solution to this problem is actually a known value thanks to the work of number theorists.

The number of ways to write 100 as a sum of at least two positive integers, which essentially means the number of partitions of 100 excluding the number 100 itself (since that is just one integer not a sum of two or more), is given as p(100) – 1 where p(n) is the partition function.

According to the computational results of partition function, p(100) equals 190,569,292. Therefore, the number of ways to write 100 as a sum of at least two positive integers would be p(100) – 1 = 190,569,291 different ways.

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