Estefânia has five cards marked with the letters $A, B, C, D$ and $E$, stacked in that order from top to bottom. She shuffles the cards by taking the top two and placing them, in reversed order, at the bottom of the stack. The figure shows what happens in the first two times she shuffles the cards.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-05.jpg?height=192&width=731&top_left_y=658&top_left_x=568)

If Estefânia shuffles the cards 74 times, which card will be on top of the stack?
A) $A$
B) $B$
C) $C$
D) $D$
E) $E$

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Solution

## ALTERNATIVE E

The reader can verify that if Estefânia shuffles the cards 6 times, they will return to their initial position. Since $74=12 \times 6+2$, shuffling the cards 74 times has the same effect as doing it twice, which leaves the card $E$ at the top of the stack.

13 Paula writes numbers - Solution

## ALTERNATIVE C

The arrow pointing downwards in the table passes through the squares of the odd numbers: $1^{2}=1,3^{2}=9$, $5^{2}=25$ and so on.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-27.jpg?height=534&width=605&top_left_y=1709&top_left_x=631)

Let's call $a_{n}$ the $n$-th term of our sequence; for example, $a_{1}=1, a_{2}=3, a_{3}=13 \text{ and } a_{4}=31$. Observing the table, we see that

$$
\begin{aligned}
& 1^{2} \xrightarrow{1 \text { space to the right }} 1^{2}+1=2 \xrightarrow{1 \text { space up }} 1^{2}+1+1=3=a_{2} \\
& 3^{2} \xrightarrow{1 \text { space to the right }} 3^{2}+1=10 \xrightarrow{3 \text { spaces up }} 3^{2}+1+3=13=a_{3} \\
& 5^{2} \xrightarrow{1 \text { space to the right }} 5^{2}+1=26 \xrightarrow{5 \text { spaces up }} 5^{2}+1+5=31=a_{4}
\end{aligned}
$$

and so on. We see, then, that the formation rule of the sequence, starting from $a_{2}$, is

$$
\begin{aligned}
& a_{2}=\left[1^{\text{st}} \text { odd }\right]^{2}+1+1^{\text{st}} \text { odd } \\
& a_{3}=\left[2^{\text{nd}} \text { odd }\right]^{2}+1+2^{\text{nd}} \text { odd } \\
& a_{4}=\left[3^{\text{rd}} \text { odd }\right]^{2}+1+3^{\text{rd}} \text { odd }
\end{aligned}
$$

and, in general,

$$
a_{n}=\left[(n-1)^{\text{th}} \text { odd }\right]^{2}+1+(n-1)^{\text{th}} \text { odd }
$$

Therefore, $a_{30}=\left[29^{\text{th}} \text { odd }\right]^{2}+1+29^{\text{th}}$ odd, and since the $29^{\text{th}}$ odd number is 57, it follows that $a_{30}=57^{2}+1+57=3307$. More generally, the $(n-1)^{\text{th}}$ odd number is $2(n-1)-1=2n-3$ and it follows that $a_{n}=(2n-3)^{2}+1+(2n-3)=$ $4n^{2}-10n+7$.

## Combinatorics

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