## Task 1 - 281211

A work collective wants to participate in the Tele-Lotto 5 out of 35 together. Colleagues $A, B, C$ are tasked with selecting the numbers on the betting slips to be submitted. During their discussion on which tips they want to compile, each of the three colleagues has certain requirements.

So, $A$ demands that each tip contains three prime numbers whose sum is 42. $B$ requires that each tip contains three numbers whose product is 33 times their sum. $C$ expects that each tip contains two numbers that are not prime numbers.

Determine all the tips that meet the requirements of all three colleagues.
I. If a tip meets the demands of all three colleagues, then:

a) According to the demands of $A$ and $C$, the tip contains exactly three prime numbers and exactly two numbers that are not prime numbers. Since all prime numbers except 2 are odd, it follows from the demand of $A$ that the tip contains the number 2.

b) Among the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, only the pairs (17, 23) and (11, 29) sum to 40. Therefore, the tip contains one of the number triples $(2 ; 17 ; 23) \quad$ (1) or $(2 ; 11 ; 29) \quad$ (2).

c) If the numbers in the demand of $B$ are denoted by $x, y$, and $z$, then $33(x+y+z)=x y z$.

Thus, one of these numbers must be divisible by 11. Without loss of generality, let this be $x$. Since $1 \leq x \leq 35$, $x$ can only take the values 11, 22, 33.

1. Case: Let $x=11$.

Then $3(11+y+z)=y z$, so $(y-3)(z-3)=42$. Since $1 \leq y \leq 35$ and $1 \leq z \leq 35$, thus $-2 \leq y-3 \leq 32$ and $-2 \leq z-3 \leq 32$, the only possibilities (up to order) are

$$
(y-3 ; z-3)=(2 ; 21),(3 ; 14),(6 ; 7)
$$

thus only

$$
(x ; y ; z)=(11 ; 5 ; 24) \quad(3), \quad(11 ; 6 ; 17) \quad(4), \quad(11 ; 9 ; 10)
$$

Now, among the five numbers of the tip, one of the triples (1), (2) and one of the triples (3), (4), (5) must be present. This is only possible with the combinations

| (1) with (4) | $(2 ; 17 ; 23 ; 11 ; 6)$ |
| :--- | :--- |
| (2) with (3) | $(2 ; 11 ; 29 ; 5 ; 24)$ |
| (2) with (4) | $(2 ; 11 ; 29 ; 6 ; 17)$ |
| (2) with (5) | $(2 ; 11 ; 29 ; 9 ; 10)$ |

Combinations (6), (7), (8) are excluded because they do not meet the demand of $C$.

2. Case: Let $x=22$.

Then $3(22+z+y)=2 y z$, so $(2 y-3)(2 z-3)=141$. Since $1 \leq y \leq 35$ and $1 \leq z \leq 35$, thus $-1 \leq 2 y-3 \leq 67$ and $-1 \leq 2 z-3 \leq 67$, the only possibility (up to order) is $(2 y-3 ; 2 z-3)=(3 ; 47)$, thus only $(x ; y ; z)=(22 ; 3 ; 25)$.

However, this cannot be combined with (1) or (2) to form five numbers of a required tip; therefore, the 2nd case is excluded.

3. Case: Let $x=33$.

Then $33+z+y=y z$, so $(y-1)(z-1)=34$. Since $0 \leq y-1 \leq 34$ and $0 \leq z-1 \leq 34$, the only possibilities (up to order) are $(y-1 ; z-1)=(1 ; 34),(2 ; 17)$, thus only $(x ; y ; z)=(33 ; 2 ; 35) \quad(10),(33 ; 3 ; 18) \quad(11)$.

These can only be combined with (1) or (2) as (10):

$$
\begin{array}{ll}
\text { (1) with }(10) & (2 ; 17 ; 23 ; 33 ; 35) \\
\text { (2) with (10) } & (2 ; 11 ; 29 ; 33 ; 35)
\end{array}
$$

d) Therefore, only the tips

$$
(2 ; 9 ; 10 ; 11 ; 29),(2 ; 11 ; 29 ; 33 ; 35),(2 ; 17 ; 23 ; 33 ; 35)
$$

can meet the demands of all three colleagues.

II. These tips meet the demands of $A$ and $C$, as they contain the prime numbers 2, 11, 29 or 2, 17, 23 with the sum of 42 and the non-prime numbers 9, 10 or 33, 35.

With I. and II., it is shown that exactly the tips specified in (14) meet the demands of all three colleagues.

## Adopted from $[5]$