# Task 4. (12 points)

Find the smallest positive integer in which the product of the digits is 5120.
# Solution.

Since $5120=2^{10} \cdot 5$, the main question is how many digits are in this number. Clearly, there are no 0s and there is a 5. From the ten factors of 2, we can form a minimum of 4 digits. This can be done in two ways: $2,8,8,8$ or $4,4,8,8$. Then, from the two sets of digits $5,2,8,8,8$ and $5,4,4,8,8$, we need to form the smallest number. The first digit should be the smallest, i.e., 2.

Answer: 25888.

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 12 |
| The idea of factoring the number into prime factors is presented. Not all factorization options are presented. The answer is correct. | $\pm$ | 9 |
| The main idea of the solution is found. All factorization options are presented. An incorrect answer is obtained. | $+/ 2$ | 6 |
|  |  |  |
| The idea of factoring the number into prime factors is presented. Not all factorization options are presented. The answer is incorrect. | $\mp$ | 2 |
| :---: | :---: | :---: |
| The answer is correct, but the solution is missing. |  |  |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 10 |