[Combinatorics]
Problem: A multiplicative partition of a positive integer $n \geq 2$ is a way of writing $n$ as a product of one or more integers, each greater than 1. Note that we consider a positive integer to be a multiplicative partition of itself. Also, the order of the factors in a partition does not matter; for example, $2 \times 3 \times 5$ and $2 \times 5 \times 3$ are considered to be the same partition of 30 . For each positive integer $n \geq 2$, define $P(n)$ to be the number of multiplicative partitions of $n$. We also define $P(1)=1$. Note that $P(40)=7$, since the multiplicative partitions of 40 are $40,2 \times 20,4 \times 10$, $5 \times 8,2 \times 2 \times 10,2 \times 4 \times 5$, and $2 \times 2 \times 2 \times 5$.

(In each part, we use "partition" to mean "multiplicative partition". We also call the numbers being multiplied together in a given partition the "parts" of the partition.)
Determine the value of $P(64)$.

Solution: We determine the multiplicative partitions of 64 by considering the number of parts in the various partitions. Note that 64 is a power of 2 so any divisor of 64 is also a power of 2 . In each partition, since the order of parts is not important, we list the parts in increasing order to make it easier to systematically find all of these.

* One part. There is one possibility: 64.
* Two parts. There are three possibilities: $64=2 \times 32=4 \times 16=8 \times 8$.
* Three parts. We start with the smallest possible first and second parts. We keep the first part fixed while adjusting the second and third parts. We then increase the first part and repeat.

We get: $64=2 \times 2 \times 16=2 \times 4 \times 8=4 \times 4 \times 4$.

* Four parts. A partition of 64 with four parts must include at least two $2 \mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With two $2 \mathrm{~s}$, the remaining two parts have a product of 16 .

We get: $64=2 \times 2 \times 2 \times 8=2 \times 2 \times 4 \times 4$.

* Five parts. A partition of 64 with five parts must include at least three $2 \mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With three $2 \mathrm{~s}$, the remaining two parts have a product of 8 .

We get: $64=2 \times 2 \times 2 \times 2 \times 4$.

$*$ Six parts. Since $64=2^{6}$, there is only one possibility: $64=2 \times 2 \times 2 \times 2 \times 2 \times 2$.

Therefore, $P(64)=1+3+3+2+1+1=11$.

Final answer: 11