Subfield: Combinatorics


Question: 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 $P(1000)$.

Solution:
First, we note that $1000=10^{3}=(2 \cdot 5)^{3}=2^{3} 5^{3}$.

We calculate the value of $P\left(p^{3} q^{3}\right)$ for two distinct prime numbers $p$ and $q$. It will turn out that this value does not depend on $p$ and $q$. This value will be the value of $P(1000)$, since 1000 has this form of prime factorization.

Let $n=p^{3} q^{3}$ for distinct prime numbers $p$ and $q$.

The integer $n$ has three prime factors equal to $p$.

In a given partition, these can be all together in one part (as $p^{3}$ ), can be split between two different parts (as $p$ and $p^{2}$ ), or can be split between three different parts (as $p, p$ and $p)$. There are no other ways to divide up three divisors of $p$.

Similarly, $n$ has three prime factors equal to $q$ which can be divided in similar ways.

We determine $P\left(p^{3} q^{3}\right)$ by considering the possible combination of the number of parts divisible by $p$ and the number of parts divisible by $q$ and counting partitions in each case. In other words, we complete the following table:

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We note that the table is symmetric, since the factors of $p$ and $q$ are interchangeable.

We proceed to consider cases, considering only those on the top left to bottom right diagonal and and those below this diagonal in the table.



Case 1: One part divisible by $p$, one part divisible by $q$

The partition must be $p^{3} q^{3}$ ( $n$ itself) or $p^{3} \times q^{3}$.

There are two partitions in this case.

Case 2: One part divisible by $p$, two parts divisible by $q$

The three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q$ and $q^{2}$.

The $p^{3}$ can occur in one of the parts divisible by $q$ or not.

This gives partitions $p^{3} \times q \times q^{2}$ and $p^{3} q \times q^{2}$ and $q \times p^{3} q^{2}$.

There are three partitions in this case. Similarly, there are three partitions with one part divisible by $q$ and two parts divisible by $p$.

Case 3: One part divisible by $p$, three parts divisible by $q$

The three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q, q$ and $q$.

The $p^{3}$ can occur in one of the parts divisible by $q$ or not.

This gives partitions $p^{3} \times q \times q \times q$ and $p^{3} q \times q \times q$.

(Note that the three divisors of $q$ are interchangeable so $p^{3}$ only needs to be placed with one of them.)

There are two partitions in this case. Similarly, there are two partitions with one part divisible by $q$ and three parts divisible by $p$.

Case 4: Two parts divisible by $p$, two parts divisible by $q$

The three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q$ and $q^{2}$.

Each of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.

If no part is a multiple of both $p$ and $q$, we have one partition: $p \times p^{2} \times q \times q^{2}$.

If one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part and two choices for which power of $q$ to include. (There is no choice for the remaining parts.) Thus, there are $2 \times 2=4$ such partitions:

$$
p^{2} q^{2} \times p \times q \quad p q^{2} \times p^{2} \times q \quad p^{2} q \times p \times q^{2} \quad p q \times p^{2} \times q^{2}
$$

If two parts are a multiple of both $p$ and $q$, there are two ways to choose the power of $p$ in the part containing just $q$, so there are two such partitions: $p q \times p^{2} q^{2}$ and $p^{2} q \times p q^{2}$. There are seven partitions in this case.

Case 5: Two parts divisible by $p$, three parts divisible by $q$

The three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q, q$ and $q$.

Each of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.

If no part is a multiple of both $p$ and $q$, we have one partition: $p \times p^{2} \times q \times q \times q$.

If one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part (since all powers of $q$ are identical).

Thus, there are 2 such partitions: $p^{2} q \times p \times q \times q$ and $p q \times p^{2} \times q \times q$.

If two parts are a multiple of both $p$ and $q$, there is one partition, since all of the powers of $q$ are identical: $p q \times p^{2} q \times q$.

There are four partitions in this case. Similarly, there are four partitions with two parts divisible by $q$ and three parts divisible by $p$.

Case 6: Three parts divisible by $p$, three parts divisible by $q$

The three factors of $p$ as $p, p$ and $p$. The three factors of $q$ appear as $q, q$ and $q$.

Here, the number of parts in the partition that are multiples of both $p$ and $q$ can be 0 , 1,2 or 3 . Since all of the powers of $p$ and $q$ are identical, the partitions are completely determined by this and are

$$
p \times p \times p \times q \times q \times q \quad p \times p \times p q \times q \times q \quad p \times p q \times p q \times q \quad p q \times p q \times p q
$$

There are four partitions in this case.



Finally, we complete the table:

Number of parts divisible by $p$ (Column)

Number of parts divisible by $q$ (Row)

|  | 1 | 2 | 3 |
| :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 2 |
| 2 | 3 | 7 | 4 |
| 3 | 2 | 4 | 4 |

Adding the entries in the table, we obtain $P\left(p^{3} q^{3}\right)=31$.

Thus, $P(1000)=31$.

Final answer: 31