10.1. If you divide 2014 by 105, the quotient is 19 and the remainder is also 19. What other natural numbers can 2014 be divided by so that the quotient and the remainder are the same? (List all possible options and prove that there are no others.)
Answer: also by 52, by 1006, and by 2013.

Solution. If 2014 is divided by a natural number $N$ and the quotient and the remainder are both natural numbers $k$, then $2014=N k+k=k(N+1)$, where $k<N$. Therefore, $k$ is a divisor of the number 2014. From the example given in the problem, it follows that $2014=105 \cdot 19+19=19 \cdot 106$, which helps factorize 2014 into prime factors: $2014=2 \cdot 19 \cdot 53$.

Now it is clear that, in addition to 19, the number 2014 has other divisors 1, 2, and 38, which generate the following division examples: $2014=1 \cdot 2014=2013 \cdot 1+1, 2014=2 \cdot 1007=1006 \cdot 2+2$, and $2014=38 \cdot 53=52 \cdot 38+38$. If the values of $k$ are taken as the other divisors of 2014 ($53, 106, 1007$, and 2014), they generate values of $N$ that do not satisfy the inequality $k<N$, so there are no other solutions.

## Grading Criteria

+ The correct answer and a fully justified solution are provided (the answer may or may not include the number 105)

$\pm$ It is proven that the remainder is a divisor of 2014, and the correct answer is provided, but it is not proven that other divisors do not yield new solutions

$\pm$ The correct reasoning is provided, but one of the possible answers is missing

干 It is proven that the remainder is a divisor of 2014, but the correct answer is not obtained

干 The correct answer is provided, and it is only shown that it satisfies the condition

- Only the answer is provided
- The problem is not solved or is solved incorrectly