## Task 1 - 340711

Armin has 100 sticks of $7 \mathrm{~cm}$ each and 100 sticks of $12 \mathrm{~cm}$ each. He wants to lay out a line segment of $1 \mathrm{~m}$ in length using such sticks. The sticks should always fit seamlessly together, and no part of the segment should be covered more than once.

Find all the possibilities for how many $7 \mathrm{~cm}$ sticks and how many $12 \mathrm{~cm}$ sticks can be combined to form such a layout!
With $x$ sticks of length $7 \text{~cm}$ and $y$ sticks of length $12 \text{~cm}$, a segment of $1 \text{~m}$ length can be laid out exactly if $7 x+12 y=100$ holds.

I. If $x$ and $y$ are two numbers that satisfy this condition, then:

Since 12 and 100 are divisible by 4, but 7 is coprime to 4, $x$ must be divisible by 4. If $x \geq 16$, then $7 x+12 y \geq 7 \cdot 16 > 100$. Therefore, $x$ can only be one of the numbers $0, 4, 8, 12$.

If $x=0$, then $12 y=100$; if $x=8$, then $12 y=100-7 \cdot 8=44$; if $x=12$, then $12 y=100-7 \cdot 16$. In all three cases, one would not get an integer for $y$.

Therefore, only $x=4$ and thus $12 y=100-7 \cdot 4=72$, so $y=6$ can be the case.

II. For $x=4$ and $y=6$, the condition is satisfied because $7 \cdot 4+12 \cdot 6=28+72=100$.

Therefore, there is exactly one way to lay out the segment using 4 sticks of $7 \text{~cm}$ and 6 sticks of $12 \text{~cm}$.