## Task 10/78

We are looking for all triples of natural numbers $(a ; b ; c)$ with the following properties:

1. None of the numbers $a ; b ; c$ is a prime number.
2. The sum $a+b+c$ lies between 1900 and 2000: $1900 \leq a+b+c \leq 2000$
3. The product $a b c$ lies between 7600 and 8000: $7600 \leq a b c \leq 8000$
4. The sum of all digits appearing in the numbers $a ; b ; c$ is 30, where no digit appears more than once in any of the numbers (which does not mean that a digit cannot appear more than once in different numbers).
5. It is $a \leq b \leq c$.
Due to condition 3, $a ; b ; c>0$. If $a \geq 4$, then due to 5., $b \geq 4$ and thus $a b c \geq 16 c$, so (due to 3.) $c \leq 500$; then, according to 5., $a ; b \leq 500$ and the sum $a+b+c \leq 1500$ would contradict 2. Therefore, only $a=1$ is possible (due to 1., $a=2$ and $a=3$ are ruled out as prime numbers). Thus, conditions 2 and 3 take the following form:

$$
1988 \leq b+c \leq 1999 \quad\left(2^{\prime}\right) \quad ; \quad 7600 \leq b c \leq 8000
$$

If $b \geq 9$, then according to 3., $c \leq 888$ and thus $b+c \leq 2 c \leq 1776$, which contradicts 2'. Therefore, $b$ can only take the values $1 ; 4 ; 6 ; 8$ (the values $2 ; 3 ; 5 ; 7$ are ruled out as prime numbers).

By substituting $b=1 ; 6 ; 8$ into 2'. and 3'. we obtain contradictions:

$b=1: 1898 \leq c \leq 1998,7600 \leq c \leq 8000$;

$b=6: 1893 \leq c \leq 1993,1266 \leq c \leq 1334$;

$b=8: 1891 \leq c \leq 1991,950 \leq c \leq 1000$

Therefore, $b=4$ must be true, which leads to $1895 \leq c \leq 1995,1900 \leq c \leq 2000$, thus $1900 \leq c \leq 1995$. For the two missing digits $x$ and $y$ of $c$, condition 4 yields the equation $1+$ $4+1+9+x+y=30$, i.e., $x+y=15$, where $x ; y \in N, x ; y \leq 9$ holds. Therefore, the numbers $c=1969, c=1978, c=1987$ are possible. The number 1969 is ruled out due to the repeated occurrence of the digit 9 (condition 4), while 1987 is ruled out as a prime number. Thus, there is only one triplet $(a . b . c)$ that satisfies all conditions: $(a . b . c)=(1.4 .1978)$.