5. Determine all natural numbers that can be represented as the product of five factors for which the following conditions hold:

- each factor is a prime number
- at least three factors are equal
- the sum of all five factors is 40.

For each such number, determine the number of its natural divisors.

## Ministry of Science and Education of the Republic of Croatia Agency for Education and Upbringing Croatian Mathematical Society

## County Competition in Mathematics February 26, 2024. 8th grade - elementary school

Each task is worth 10 points. In addition to the final result, the procedure is also scored. To earn all points, it is necessary to find all solutions and confirm that there are no others, write down the procedure, and justify your conclusions.
5. Determine all natural numbers that can be represented as the product of five factors for which the following conditions hold:

- each of the factors is a prime number
- at least three factors are equal
- the sum of all five factors is 40.

For each such number, determine the number of its natural divisors.

## Solution.

If all five factors are equal, we have $5a = 40$ or $a = 8$, which is not a prime number, so the condition of the problem is not satisfied. If four factors are equal, we have $4a + b = 40$, so $b$ is an even number, i.e., $b = 2$. This gives $4a = 38$, or $a = 9.5$, which also does not satisfy the condition of the problem.

1 POINT

Let $a, b$, and $c$ be the prime factors of the desired number, where $a$ is the prime factor that appears three times. The remaining two factors may or may not be equal.

We have $3a + b + c = 40$. The following possibilities exist:

| $a$ | $3a$ | $b+c$ | $a^3 \cdot b \cdot c$ |
| :---: | :---: | :---: | :---: |
| 2 | 6 | 34 | $\mathbf{2}^{3} \cdot \mathbf{3} \cdot \mathbf{31}=\mathbf{744}$ |
| 2 | 6 | 34 | $\mathbf{2}^{3} \cdot \mathbf{5} \cdot \mathbf{29}=\mathbf{1160}$ |
| 2 | 6 | 34 | $\mathbf{2}^{3} \cdot \mathbf{11} \cdot \mathbf{23}=\mathbf{2024}$ |
| 2 | 6 | 34 | $\mathbf{2}^{3} \cdot \mathbf{17} \cdot \mathbf{17}=\mathbf{2312}$ |
| 3 | 9 | 31 | $\mathbf{3}^{3} \cdot \mathbf{2} \cdot \mathbf{29}=\mathbf{1566}$ |
| 5 | 15 | 25 | $\mathbf{5}^{3} \cdot \mathbf{2} \cdot \mathbf{23}=\mathbf{5750}$ |
| 7 | 21 | 19 | $\mathbf{7}^{3} \cdot \mathbf{2} \cdot \mathbf{17}=\mathbf{11662}$ |
| 11 | 33 | 7 | $\mathbf{11}^{3} \cdot \mathbf{2} \cdot \mathbf{5}=\mathbf{13310}$ |
| 13 | 39 | 1 | no solution |

If the factorization includes three prime numbers greater than or equal to 13, the sum of the remaining two factors will be less than 2, so we conclude that the solutions listed are the only ones.

1 POINT

If $x$ is the desired number, with the prime factorization $x = a^3 \cdot b \cdot c$, then the divisors of $x$ are the numbers $1, a, a^2, a^3, b, c, ab, bc, ac, a^2b, a^2c, a^3b, a^3c, abc, a^2bc, a^3bc$, and there are 16 of them.

3 POINTS

Similarly, if $x$ is the desired number, with the prime factorization $x = a^3 \cdot b^2$, then the divisors of $x$ are the numbers $1, a, a^2, a^3, b, ab, b^2, a^2b, a^3b, ab^2, a^2b^2, a^3b^2$, and there are 12 of them.

1 POINT

Therefore, the numbers 744, 1160, 2024, 1566, 5750, 11662, and 13310 each have 16 divisors, while the number 2312 has 12 divisors.

Note 1. The number of divisors can also be determined combinatorially as follows.

Each divisor of the product $a^3bc$ is of the form $a^m \cdot b^n \cdot c^k$ for $m \in \{0,1,2,3\}, n \in \{0,1\}, k \in \{0,1\}$, so the number of natural divisors of such a number is $4 \cdot 2 \cdot 2 = 16$.

Similarly, the divisors of the product $a^3b^2$ are of the form $a^m \cdot b^n$, where $m \in \{0,1,2,3\}, n \in \{0,1,2\}$, so the number of natural divisors of such a number is $4 \cdot 3 = 12$.

In both cases, the same reasoning is used, and it does not matter in which order the cases are considered. If the student considers only one case and reaches the correct conclusion, they should receive 3 POINTS.

This grading guideline applies to the approach presented in the solution.

Note 2. The student can determine the number of divisors by listing all divisors.

| $a^3bc$ | Divisors | Total number of divisors |
| :---: | :---: | :---: |
| 744 | $1,2,3,4,6,8,12,24,31,62,93,124,186,248,372,744$ | 16 |
| 1160 | $1,2,4,5,8,10,20,29,40,58,116,145,232,290,580,1160$ | 16 |
| 2024 | $1,2,4,8,11,22,23,44,46,88,92,184,253,506,1012,2024$ | 16 |
| 2312 | $1,2,4,8,17,34,68,136,289,578,1156,2312$ | 12 |
| 1566 | $1,2,3,6,9,18,27,29,54,58,87,174,261,522,783,1566$ | 16 |
| 5750 | $1,2,5,10,23,25,46,50,115,125,230,250,575,1150,2875,5750$ | 16 |
| 11662 | $1,2,7,14,17,34,49,98,119,238,343,686,833,1666,5831,11662$ | 16 |
| 13310 | $1,2,5,10,11,22,55,110,121,242,605,1210,1331,2662,6655,13310$ | 16 |

1 POINT should be awarded for each two rows of the table.