1. When dividing a certain natural number by 19 and 99, the remainders are two prime numbers. The sum of both incomplete quotients equals 1999. Determine the number being divided.
SOLUTION. Both divisions of the sought number $N$ can be expressed by the equations

$$
N=19 a+p \quad \text { and } \quad N=99 b+q
$$

where $a, b$ are the respective incomplete quotients and $p, q$ are the respective remainders. According to the problem, the numbers $p, q$ are prime numbers, and as remainders, they satisfy the inequalities $p<19$ and $q<99$. The non-negative integers $a, b$ are such that their sum equals 1999. Therefore, $b=1999-a$ and from the two expressions for the number $N$,

$$
N=19 a+p=99 \cdot(1999-a)+q,
$$

we derive the equation $118 a+(p-q)=197901$. Since the difference of the remainders $p-q$ is a "small" number, specifically $-99<p-q<19$, according to the last equation, the number $118 a$ is such a multiple of the number 118 that lies between the numbers 197901 - 19 and $197901+99$. By dividing 197901 : 118, we find that $197901=1677 \cdot 118+15$. Therefore, it must be that $a=1677$ (so $b=322$) and $p-q=15$. From the last equation, it follows that one of the prime numbers $p, q$ is odd and the other is even, so $q=2$ and $p=17$. It remains to calculate the value of $N$:

$$
N=19 \cdot 1677+17=99 \cdot 322+2=31880 \text { . }
$$

Note. Some solvers will proceed by first estimating the size of the sought number $N$, for example by solving the "approximate" equation $\frac{N}{19}+\frac{N}{99}=$ $=1999$, which has the root $N \doteq 31865$. Then, using the nearest multiples of the numbers 19 and 99 ($19 \cdot 1677=31863, 99 \cdot 322=31878, 19 \cdot 1678=31882$), they will determine the only possible values of the incomplete quotients $a, b$ (it is sufficient to discuss the cases $N<31878$, $31878 \leqq N<31882$ and $N \geqq 31882$).

Answer: The sought number is 31880.