## Task A-1.7.

Let $1=d_{1}<d_{2}<d_{3}<d_{4}<d_{5}<d_{6}=n$ be all the natural divisors of the number $n$ such that $d_{5}=289$ and $d_{3}-d_{2}=10$. Determine $n$.
## Second Solution.

Since $d_{5}=289$ is a divisor of $n$, and 17 is its divisor, we conclude that 17 is also a divisor of the number $n$.

Since $d_{1}=1<17<17^{2}=d_{5}$, we have three possibilities: $d_{2}=17, d_{3}=17$ or $d_{4}=17$. \quad 1 point

If $d_{2}=17$, then $d_{3}=10+d_{2}=27$. However, in this case $3 \mid 27$ and $d_{3}=27 \mid n$, so 3 is also a divisor of $n$. It is smaller than 17, which contradicts $d_{2}=17$.

If $d_{3}=17$, then $d_{2}=d_{3}-10=7$. In this case, all numbers

$$
1,7,17,7 \cdot 17,17^{2}, 7 \cdot 17^{2}
$$

are divisors of $n$. There are 6 of them, which means there cannot be more, which is only possible if $n$ is equal to the last of them: $n=7 \cdot 17^{2}=2023$.

1 point

By direct verification, we see that the number $17^{2} \cdot 7=2023$ indeed has 6 divisors (which we also see from above), and that 289 is its fifth largest divisor.

The remaining case is $d_{4}=17$.

Note that $d_{2}$ is relatively prime to 17 since it is smaller than it. 1 point

Since $d_{2} \mid n$ and $17 \mid n$, we conclude that $17 d_{2} \mid n$, i.e., the number $17 d_{2}$ is a divisor of $n$.

Thus, there exists $i \in\{1,2,3,4,5,6\}$ such that $d_{i}=17 d_{2}$, and all divisors are ordered by size. On the other hand, we have

$$
d_{4}=17<17 d_{2}<17^{2}=289=d_{5}
$$

which leads to a contradiction.

Therefore, $n=17^{2} \cdot 7=2023$ is the only number with the desired property.

Note: A solution that does not verify that $n=2023$ satisfies all conditions of the problem is worth a maximum of 9 points (it does not achieve the tenth point of the first scoring scheme, or the fifth point of the second scoring scheme). Finding the number $n=2023$ without verifying all conditions of the problem earns 0 points (the same point is not achieved). Finding the number $n=2023$ and verifying all conditions of the problem earns the above-mentioned 1 point.

Verifying that the possibility $d_{3}=17$ leads to a potential solution $n=2023$ earns 1 point. Proving that the case $d_{2}=17$ does not lead to a solution earns 1 point.

The claims $n=d_{2} d_{5}$ and $n=d_{3} d_{4}$ can be used for parts of the proof. Therefore, solutions that include at least one of these two equalities and earn less than or equal to 1 point in total, achieve an additional 1 point.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - A variant

January 26, 2023.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO SCORE AND EVALUATE THAT APPROACH APPROPRIATELY.