2.17. Final round of the "Rosatom" Olympiad, 7th grade

# Answers and solutions 

Problem 1 Answer: 9

There exists a set of 8 buttons in which there are no three buttons of the same color: each color has two buttons. In any set of 9 buttons, there will be at least one triplet of buttons of the same color.

If we assume the opposite, that there are no more than 2 buttons of the same color, then there can be no more than 8 such buttons in total, which contradicts the condition.
Answer: 9 buttons.

Problem 2 Answer: 1261

\[
\left\{\begin{array}{l}
a=35 n+1 \\
a=45 m+1
\end{array} \rightarrow 35 n=45 m \rightarrow 7 n=9 m \rightarrow\left\{\begin{array}{c}
n=9 t \\
m=7 t, t \in Z
\end{array} \rightarrow\right.\right.
\]

\[
a=315 t+1 \geq 1000 \rightarrow t \geq 4 \rightarrow a_{\text {min }}=1261
\]

## Problem 3 Answer: 1) 4026 2) Sasha

At each step of the game, the sum of the numbers written on the cards in the bag does not change.

At the beginning of the game, it was \(2013 \times 2=4026\). Each step of the game changes the parity of the number of cards in the bag, as it reduces their number by one. Initially, the total number of cards in the bag is odd, so Sasha will make his move when the number of cards in the bag is odd, and Dasha when it is even. The last move occurs when there is one card in the bag, i.e., their number is odd, and Sasha will notice this.

Problem 4 Answer: 1) \(17 \quad\) 2) 90

\((12,18,21,24,27,30,36,42,45,48,54,60,63,72,81,84,90)\)

\(x, y\) - digits of the desired number

\[
a=10 x+y=3 k, \quad a=10 x+y=(x+y) m \rightarrow\left\{\begin{array}{c}
x+y=\frac{3 k}{m}, 2 \leq m \leq 10, \\
10 x+y=3 k, 4 \leq k \leq 33
\end{array}\right.
\]

\[
\left\{\begin{array}{l}
x=\frac{k}{3} \cdot \frac{m-1}{m} \\
y=\frac{k}{3} \cdot \frac{10-m}{m}
\end{array}\right.
\]

Case \(m=2 \rightarrow\left\{\begin{array}{l}
x=k / 6 \\
y=4 k / 3
\end{array} \rightarrow k=6 t \rightarrow\left\{\begin{array}{l}
x=t \\
y=8 t
\end{array} \rightarrow t=1 \rightarrow a=18\right.\right.
\]

Case \(m=3 \rightarrow\left\{\begin{array}{l}
x=2 k / 9 \\
y=7 k / 9
\end{array} \rightarrow k=9 t \rightarrow\left\{\begin{array}{l}
x=2 t \\
y=7 t
\end{array} \rightarrow a=27\right.\right.
\]

Case \(m=4 \rightarrow\left\{\begin{array}{l}
x=k / 4 \\
y=k / 2
\end{array} \rightarrow k=4 t \rightarrow\left\{\begin{array}{l}
x=t \\
y=2 t
\end{array} \rightarrow a=12,24,36,48\right.\right.
\]

Case \(m=5 \rightarrow\left\{\begin{array}{c}
x=4 k / 15 \\
y=k / 3
\end{array} \rightarrow k=15 t \rightarrow\left\{\begin{array}{l}
x=4 t \\
y=5 t
\end{array} \rightarrow a=45\right.\right.
\]

Case \(m=6 \rightarrow\left\{\begin{array}{l}
x=5 k / 18 \\
y=2 k / 9
\end{array} \rightarrow k=18 t \rightarrow\left\{\begin{array}{l}
x=5 t \\
y=4 t
\end{array} \rightarrow a=54\right.\right.
\]

Case \(m=7 \rightarrow\left\{\begin{array}{l}
x=2 k / 7 \\
y=k / 7
\end{array} \rightarrow k=7 t \rightarrow\left\{\begin{array}{l}
x=2 t \\
y=t
\end{array} \rightarrow a=21,42,63,84\right.\right.
\]

Case \(m=8 \rightarrow\left\{\begin{array}{c}
x=7 k / 24 \\
y=k / 12
\end{array} \rightarrow k=24 t \rightarrow\left\{\begin{array}{l}
x=7 t \\
y=2 t
\end{array} \rightarrow a=72\right.\right.
\]

Case \(m=9 \rightarrow\left\{\begin{array}{c}
x=8 k / 27 \\
y=k / 27
\end{array} \rightarrow k=27 t \rightarrow\left\{\begin{array}{c}
x=8 t \\
y=t
\end{array} \rightarrow a=81\right.\right.
\]

Case \(m=10 \rightarrow\left\{\begin{array}{c}
x=3 k / 10 \\
y=0
\end{array} \rightarrow k=10 t \rightarrow\left\{\begin{array}{l}
x=3 t \\
y=0
\end{array} \rightarrow a=30,60,90\right.\right.
\]

\[
a_{\max }=90, \text{ number of numbers } 17
\]

Problem 5 Answer: 2 weights

Indicate the minimum number of 3 kg weights needed to weigh loads of 1, 2, 3, and 4 kg.

Load of 1 kg - 2 weights of 3 kg are needed

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=198&width=505&top_left_y=438&top_left_x=633)

Load of 2 kg or 3 kg - 1 weight of 3 kg is needed

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=186&width=1076&top_left_y=796&top_left_x=229)

Load of 4 kg - 2 weights of 3 kg are needed

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=246&width=497&top_left_y=1105&top_left_x=606)

The weight of any load when divided by 5 has a remainder of 0, 1, 2, 3, or 4. If the remainder is 0, the load can be weighed using only 5 kg weights. If the weight of the load is \(5 n+1\), the weighing picture is obtained from the picture of weighing 1 kg:

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=291&width=563&top_left_y=1577&top_left_x=701)

If the weight of the load is \(5 n+2\), the weighing picture can be used for a 2 kg load:

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=220&width=514&top_left_y=1964&top_left_x=674)

If the weight of the load is \(5 n+3\), the weighing picture can be used for a 3 kg load:

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-3.jpg?height=196&width=507&top_left_y=2373&top_left_x=703)

Finally, a load of \(5 n+4\) kg is weighed similarly to a 4 kg load:

![](https://cdn.mathpix.com/cropped/2024_05_06_131aab484f98747a9500g-4.jpg?height=274&width=508&top_left_y=149&top_left_x=686)