## Task 3 - 340713

Franziska is looking for a four-digit natural number $z$ that satisfies the following statements (1), (2), and (3):

(1) The units digit of $z$ is 1 greater than the tens digit of $z$.

(2) The hundreds digit of $z$ is twice the tens digit of $z$.

(3) The number $z$ is twice a prime number.

Prove that there is exactly one such number; determine this number!
I. If statements (1), (2), and (3) hold for a natural number $z$, then:

From (3), $z$ is even, so it has an even units digit.

From (1), the tens digit is therefore odd. By doubling it, according to (2), the hundreds digit is formed, which is a number less than 10. Thus, the tens digit is less than 5 and can only be one of the numbers $1; 3$.

From (1) and (2), it follows that: The last three digits of $z$ can only be either 212 or 634.

If they were 212, the number formed by the last two digits of $z$ would be divisible by 4; thus, $z$ would be divisible by 4 and therefore (and because $z>4$) not double a prime number. Therefore, the last three digits of $z$ can only be 634.

The first digit of $z$ cannot be any of the digits 2, 5, 8; because in these cases, $z$ would have a sum of digits divisible by 3 and would therefore be divisible by 3, and thus (because $z>6$) could not be double a prime number. If the first digit of $z$ were one of the numbers $1, 3, 4, 6, 7$, then statement (3) would also not hold, as the following calculations show:

| $z$ | Factorization of $z: 2$ |
| :---: | :--- |
| 1634 | $817=19 \cdot 43$ |
| 3634 | $1817=23 \cdot 79$ |
| 4634 | $2317=7 \cdot 331$ |
| 6634 | $3317=31 \cdot 107$ |
| 7634 | $3817=11 \cdot 347$ |

Thus, the only possibility left is $z=9634$.

II. For $z=9634$, (1) and (2) are satisfied, and (3) also holds because $z: 2=4817$ is a prime number, which can be shown as follows:

Since $4817<4900=70^{2}$, if 4817 were composite, it would have a prime factor less than 70. However, it can be verified that 4817 is not divisible by any of the prime numbers 2, 3, 5, 7, 11, 13, 17, $19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67$.

With I. and II., it has been proven that there is exactly one four-digit natural number for which (1), (2), and (3) hold, and this number is determined to be $z=9634$.

## Solution adopted from $[5]$