## Task 4 - 260824

a) Determine all two-digit natural numbers $z$ that satisfy the following condition:

If you place a third digit in front of the two digits of $z$, a three-digit number is formed that is 29 times as large as $z$.

b) State how further natural numbers $z^{\prime}$ can be formed that satisfy the following condition:

If you place an additional digit in front of all the digits of $z^{\prime}$, a new number is formed that is 29 times as large as $z^{\prime}$.

c) Determine all natural, non-multiples of 10, numbers $z^{\prime \prime}$ that satisfy the following condition:

If you place an additional digit or multiple additional digits in front of all the digits of $z^{\prime \prime}$, a new number is formed that is 29 times as large as $z^{\prime \prime}$.
a) If $z$ together with a preceding digit $a$ satisfies the conditions of the problem, then $100a + z = 29z$, so $z = \frac{25a}{7}$ (1),

furthermore, $z$ is a natural number, so $25a$ must be divisible by 7. Since 25 is coprime with 7, $a$ must be divisible by 7. Additionally, $a \neq 0$ (because if $a=0$, then $z=0$, which contradicts the two-digit nature of $z$). The only non-zero digit divisible by 7 is $a=7$.

Thus, from (1), it follows that only $z=25$ can satisfy the conditions of the problem, and this is together with the preceding digit $a=7$.

Indeed, these conditions are satisfied, because $725 = 29 \cdot 25$.

b) Numbers $z'$ with the required property can be formed, for example, by appending any number of zeros to the number 25 found above; because $72500 \ldots 0 = 29 \cdot 2500 \ldots 0$.

c) If $z''$ together with a preceding sequence of digits, which represents the natural number $b$, satisfies the conditions of the problem and if $z''$ is an $n$-digit natural number, then

$$
10^n \cdot b + z'' = 29z'' \quad \Rightarrow \quad z'' = \frac{10^n \cdot b}{28}
$$

Furthermore, $z''$ is a natural number, so $10^n \cdot b$ must be divisible by (28 and thus by) 7. Since $10^n$ is coprime with 7, $b$ must be divisible by 7. Additionally, $b \neq 0$; because if $b=0$, then $z''=0$, which contradicts the condition in the problem that $z''$ should not be divisible by 10.

From the fact that $z''$ is an $n$-digit number, it follows that $z'' < 10^n$; from this and (2), it follows that $\frac{b}{28} < 1$, so $b < 28$.

Thus, $b$ must be one of the numbers 7, 14, 21. In these three cases, we get

$$
z'' = \frac{10^n}{4} = 10^{n-1} \cdot 25 \quad \text{or} \quad z'' = \frac{10^n}{2} = 10^{n-1} \cdot 5 \quad \text{or} \quad z'' = \frac{10^n \cdot 3}{4} = 10^{n-2} \cdot 75
$$

This is only a natural number not divisible by 10 if $n=2$ or $n=1$ or $n=2$, respectively. This leads to $z'' = 25$ or $z'' = 5$ or $z'' = 75$.

Thus, only these numbers can satisfy the conditions of the problem, and only together with the preceding sequence of digits $b=7$ or $b=14$ or $b=21$.

Indeed, these conditions are satisfied, because the mentioned numbers $z''$ are not divisible by 10, and it holds that

$$
725 = 29 \cdot 25 \text{ or } 145 = 29 \cdot 5 \text{ or } 2175 = 29 \cdot 75
$$

Solutions of the 2nd Round 1986 adopted from [5]

### 5.28.3 3rd Round 1986, Class 8