Task 2. (10 points)

Find the denominator of the fraction $\frac{100!}{28^{20}}$ after it has been reduced to its simplest form.

(The expression 100! is equal to the product of the first 100 natural numbers: $100!=1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100$.$) .$
# Solution.

It can be noticed that among the numbers from 1 to 100, exactly 14 numbers are divisible by 7, and exactly two are divisible by 49. Therefore, in the prime factorization of $1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100$, the number 7 will appear in the power of 16. Among the numbers from 1 to 100, exactly 50 are even, so in the prime factorization of $1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100$, the number 2 will appear in a power no less than 50. Therefore, in the fraction $\frac{100!}{28^{20}}=\frac{100!}{7^{20} \cdot 2^{40}}$, after simplification, only $7^{4}=2401$ will remain in the denominator.

Answer: 2401.

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 10 |
| A minor computational error was made. | $\pm$ | 7 |
| The main idea of the solution was found. The solution is incomplete. The answer is correct. | $+/ 2$ | 5 |
| The main idea of the solution was found. The solution is incomplete. The answer is incorrect. | $\mp$ | 2 |
| The solution does not meet any of the criteria described above. | $-/ 0$ | 0 |
| Maximum score |  | 10 |