Task 1. Find the smallest natural number that has the following property: the remainder of its division by 20 is one less than the remainder of its division by 21, and the remainder of its division by 22 is 2.
Answer: 838.

Solution. The desired number is $20k + a = 21l + a + 1 = 22m + 2$, where $0 \leqslant a \leqslant 19$ and $l, k, m \geqslant 0$. From the first equation and the congruence modulo 20, we get that $l + 1 \equiv 0 \pmod{20}$. Since we are looking for the smallest number, let's try $l = 19$, if this $l$ does not work, then consider $l = 19 + 20 = 39$, and so on. We seek the desired number in the form $21 \cdot 19 + a + 1 = 22m + 2$, compare modulo 22: $3 + a + 1 \equiv 2 \pmod{22}$, hence, $a \equiv 20 \pmod{22}$, which is impossible. Now we seek the desired number in the form $21 \cdot 39 + a + 1 = 22m + 2$, so $5 + a + 1 \equiv 2 \pmod{22}$, that is, $a \equiv 18 \pmod{22}$. The number $21 \cdot 39 + 18 + 1 = 838$ satisfies the condition and is the smallest by construction.