What is the smallest natural number, for which, omitting its first digit, the resulting number is four times a prime number, and the number following the resulting number is five times a prime number?
The first digit of the number in question is clearly 1, because in any other case, by writing 1 in the place of the first digit, we would get a smaller number with the same property.

Let the number formed by omitting the first digit be $N$. According to the condition, $N=4p$ and $N+1=5q$, where $p$ and $q$ are prime numbers. From this,

$$
q=5q-4q=4p+1-4q=4(p-q)+1=4k+1
$$

and

$$
4p=5q-1=20k+4, \quad \text{that is} \quad p=5k+1
$$

where $k$ is an integer. As $k$ increases, $p$, and thus $N$, also increases, so our task is to determine the smallest integer $k$ for which $p=5k+1$ and $q=4k+1$ are both prime numbers.

Since $N$ is non-negative, $p$ and $q$ are also non-negative, and since they are primes, they are greater than 1, so $k \geq 1$. $k$ must be even, because otherwise $5k+1$ would be an even composite number. $k$ must also be divisible by 3, because if it left a remainder of 1, then $5k+1$ would be divisible by 3 and greater than 3, thus composite, and if it left a remainder of 2, then $4k+1$. Therefore, $k$ is divisible by 2 and 3, hence by 6: $k=6n$, $p=30n+1$, and $q=24n+1$, and our task is to find the smallest positive integer $n$ for which both $p$ and $q$ are prime.

By trying the first few values, we have listed in the table below the prime factorization of each composite value:

$$
\begin{aligned}
& n=1 \quad \text{in which case} \quad 24n+1=5^{2} \\
& n=2 \quad \text{in which case} \quad 24n+1=7^{2} \\
& n=3 \quad \text{in which case} \quad 30n+1=7 \cdot 13 \\
& n=4 \quad \text{in which case} \quad 30n+1=11^{2}, \\
& n=5 \quad \text{in which case} \quad 24n+1=11^{2} \\
& n=6 \quad \text{in which case} \quad 24n+1=5 \cdot 29, \\
& n=7 \quad \text{in which case} \quad 24n+1=13^{2} \\
& n=8 \quad \text{in which case} \quad 30n+1=241, \quad 24n+1=193 \text{. }
\end{aligned}
$$

It is easy to verify that 241 and 193 are both prime. Thus, $N=4 \cdot 241=964$, and the number we are looking for is 1964.