Two prime numbers differ by a hundred. When their decimal representations are written one after the other, we get another prime number. Which are these numbers?
Let the two prime numbers be denoted by $p$ and $q$, with $p < q$, and let the prime number formed by writing them consecutively be denoted by $r$.

Let's examine the remainder that $p$ can give when divided by 3. For this, we use the known fact that a positive integer gives the same remainder when divided by 3 as the sum of its digits in the decimal system.

If $p$ gives a remainder of 1 when divided by 3, then $q = p + 100$ clearly gives a remainder of 2, because both $p$ and 100 give a remainder of 1. In this case, however, $r$, whose digits are given by $p$ and $q$, is divisible by 3, since the sum of the digits of $p$ gives 1 and the sum of the digits of $q$ gives 2. The at least four-digit $r$ is therefore divisible by 3 and cannot be prime.

If $p$ gives a remainder of 2 when divided by 3, then $q = p + 100$ is divisible by 3, but it cannot be 3 itself, as it is at least three digits. In this case, therefore, $q$ is not prime.

If $p$ is divisible by 3, then only $p = 3$ is possible. From this, $q = p + 100 = 103$ (which is prime) follows. Writing these two numbers consecutively, we get the numbers 3103 and 1033. Of these, $3103 = 29 \cdot 107$ is composite, while 1033 is prime.

The only solution to the problem is therefore: $p = 3, q = 103, r = 1033$.

Based on the work of Csaba Megyeri (Nagykanizsa, Batthyány L. Gymnasium, 2nd grade)