## Task 4 - 180924

Determine all three-digit natural numbers $z$ that have the following properties (1) to (4):

(1) $z$ is a prime number.

(2) Each digit of $z$ represents a prime number.

(3) The cross sum $z^{\prime}$ of $z$ is a two-digit prime number.

(4) The cross sum $z^{\prime \prime}$ of $z^{\prime}$ is a prime number.
Assuming a three-digit number $z$ has the properties (1) to (4).

Due to (2), only the following digits can appear in it: 2, 3, 5, and 7.

Because of (1), the digits 2 and 5 cannot appear as the last digit. Therefore, $z$ must end with the digit 3 or 7.

Since the cross sum $z'$ is a two-digit prime number, which is formed as the sum of three summands, none of which is greater than 7, $z'$ can only be one of the numbers 11, 13, 17, or 19. Of these, only $z' = 11$ has a prime number, namely the number 2, as its cross sum.

Thus, $z' = 11$.

Now, let the last digit of $z$ be the number 7. Then the sum of the numbers represented by the first two digits must be 4. Of the possible decompositions of the number 4 into two natural numbers as summands (namely $0+4 ; 1+3 ; 2+2 ; 3+1$ and $4+0$), only $2+2$ satisfies condition (2). This results in $z = 227$.

Now, let 3 be the last digit of $z$. Then the sum of the numbers represented by the first two digits of $z$ must be 8.

Of the possible decompositions of the number 8 into two natural numbers as summands (namely $0+8$; $1+7 ; 2+6 ; 3+5 ; 4+4 ; 5+3 ; 6+2 ; 7+1 ; 8+0$), only $3+5$ and $5+3$ satisfy condition (2).

This leads to the numbers $z = 353$ and $z = 533$.

Since $533 = 13 \cdot 41$, the number 533 does not satisfy condition (1).

Therefore, the numbers 227 and 353 can at most satisfy the conditions of the problem.

They actually do; because 227 and 353 are prime numbers. (Proof: 227 is not divisible by any prime number $p < \sqrt{227}$, 353 is not divisible by any prime number $p < \sqrt{353}$.) Their digits 2, 2, 7 and 3, 5, 3 are also prime numbers. The same is true for their cross sum 11. Finally, the cross sum 2 of 11 is a prime number, as required.

Solutions of the 2nd Round 1978 adopted from [5]

### 6.20.3 3rd Round 1978, Class 9