## Task 36/69

We are looking for a natural, four-digit number with the following properties:

1. The sum of the thousands place and the hundreds place is equal to the number that results when the two middle digits are removed from the sought number.
2. This sum is less than double the tens place.
3. Exactly one of the four place values is a prime number.
Let $a, b, c$ and $d$ be the thousands, hundreds, tens, and units place of the number $z$ we are looking for. Then we have

$$
z=1000 a+100 b+10 c+d
$$

with $1 \leq a \leq 9, 0 \leq b ; c ; d \leq 9, a, b, c, d$ being integers (we can exclude $a=0$ because otherwise $z$ would not be a true four-digit number). From condition 1, we get

$$
a+b=10 a+d \quad \text { or } \quad b=9 a+d
$$

From this, it immediately follows that $a=1, b=9, d=0$. Since none of these three numbers are prime, $c=p$ must be a prime number according to condition 3. From condition 2, we now get

$$
a+b<2 c=2 p ; \quad 10<2 c=2 \beta ; \quad 5<c=p
$$

Thus, $c=p=7$. Therefore, $z=1970$.

## Solution by Nguyenthitrong Hien Tarrago:

Let the number we are looking for be $1000 a+100 b+10 c+d$. Then we have

$$
a+b=10 a+d \quad ; \quad a+b<2 c
$$

The number $c$ is not greater than 9: $c \leq 9$, so $a+b<18, 10+d \leq 18$.

From this, it follows that $a=1, d \leq 8, b=9+d$. The number $b$ is not greater than 9: $b \leq 9$. Therefore, $b=9$ and $d=0$. Now we have

$$
1+9<2 c \leq 18 \quad ; \quad 5<c \leq 9
$$

Thus, $c=7$ (since $c$ must be a prime number, and 7 is the only prime number between 5 and 9).

The number we are looking for is 1970.

### 2.10 Problems and Solutions 1970