Problem 6.8. A natural number $n$ is called good if 2020 gives a remainder of 22 when divided by $n$. How many good numbers exist?
Answer: 10.

Solution. In the subsequent solution, the expression of the form $a^{k}$ - the number $a$ raised to the power of $k$ - is the number $a$ multiplied by itself $k$ times. We will also consider $a^{0}=1$. For example, $3^{2}=3 \cdot 3=9$, and $2^{0}=1$.

Since 2020 gives a remainder of 22 when divided by $n$, then $2020-22=1998$ is divisible by $n$, and also $n>22$ (since the remainder of the division is less than the divisor). Therefore, we need to find the number of divisors of 1998 that are greater than 22. Clearly, any such number is a good number.

Let's factorize 1998 into prime factors: $1998=2 \cdot 3^{3} \cdot 37$. Any of its divisors that are divisible by 37 are greater than 22, so they are good numbers. Such divisors are represented in the form $2^{a} \cdot 3^{b} \cdot 37$, where $a$ can take one of the values 0,1 (2 options), and $b$ can take one of the values $0,1,2,3$ (4 options). Therefore, there are $2 \cdot 4=8$ such divisors.

Now let's count the number of divisors that are not divisible by 37. Such divisors are represented in the form $2^{c} \cdot 3^{d}$, where $c$ can take one of the values 0,1, and $d$ can take one of the values $0,1,2,3$. If $d>22$, so it is a good number. There are only two such divisors: $3^{3}$ and $2 \cdot 3^{3}$.

In total, there are $8+2=10$ good numbers.

Remark. Knowing that $1998=2 \cdot 3 \cdot 3 \cdot 3 \cdot 37$, we can simply list all 16 divisors of this number and verify that exactly 10 of them exceed 22:

$$
1,2,3,6,9,18,27,37,54,74,111,222,333,666,999,1998
$$

## 7th grade