# 2021 AIME II Problems/Problem 3

## Problem

Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products $$x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$$ is divisible by $3$.

## Solution 1

Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3$, so WLOG $x_3=3$, we will multiply by $5$ afterward since any of $x_1, x_2, ..., x_5$ would be $3$, after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3$, since $x_5x_1$ is never divisible by $3$, now we just need to find the number of ways $x_4+x_2$ is divisible by $3$, after some calculation you will see that there are $16$ ways to choose $x_1, x_2, x_4,$ and $x_5$ in this way. So the desired answer is $16 \times 5=\boxed{080}$.

~ math31415926535

## Solution 2 (Cyclic Symmetry and Casework)

The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.$ We have:

1. $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\equiv x_2x_3x_4 + x_3x_4x_5\pmod{3}.$
2. $x_2,x_3,x_4,x_5$ are congruent to $1,2,1,2\pmod{3}$ in some order.

We construct the following table for the case $x_1=3,$ with all values in modulo $3:$ $$\begin{array}{c||c|c|c|c|c||c} & & & & & & \\ [-2.5ex] \textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\ 2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\ 3 & 1 & 2 & 2 & 1 & 2 & \\ 4 & 2 & 1 & 1 & 2 & 1 & \\ 5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\ 6 & 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}$$ For Row 1, $(x_2,x_3)$ can be either $(1,4)$ or $(4,1),$ and $(x_4,x_5)$ can be either $(2,5)$ or $(5,2).$ By the Multiplication Principle, Row 1 produces $2\cdot2=4$ permutations. Similarly, Rows 2, 5, and 6 each produce $4$ permutations.

Together, we get $4\cdot4=16$ permutations for the case $x_1=3.$ By the cyclic symmetry, the cases $x_2=3, x_3=3, x_4=3,$ and $x_5=3$ all have the same count. Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\cdot5=\boxed{080}.$

~MRENTHUSIASM

## Solution 3

WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$.

We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$. We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 (mod 3)$ The only way is when They are $(+1,-1)$ or $(-1, +1)$ (mod 3)

The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1$ (mod $3$) respectively. $+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= +1 \cdot +1 \cdot +1$ $\;\;\; OR \;\;\;+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \cdot -1 \cdot +1$. $-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= -1 \cdot -1 \cdot -1$ $\;\;\; OR \;\;\;-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \cdot +1 \cdot +1$

But, as we have just two $+1's$ and two $-1's$, Hence, We will have to take $+1 = +1 \cdot -1 \cdot -1$ and $-1 = -1 \cdot +1 \cdot +1$ Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$ are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ )

So, $(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values.

For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$ Hence, in total, we have $8 \times 2 = 16$ ways.

But any of the $x_{i} 's$ can be $3$ So, $16 \times 5 = \boxed{080}$

-Arnav Nigam

## See Also

 2021 AIME II (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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