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Difference between revisions of "2021 AIME II Problems/Problem 3"

(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
 
The expression <math>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</math> has cyclic symmetry. Without the loss of generality, let <math>x_1=3.</math> It follows that <math>\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.</math> We have  
 
The expression <math>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</math> has cyclic symmetry. Without the loss of generality, let <math>x_1=3.</math> It follows that <math>\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.</math> We have  
−
# <math>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}</math>
+
# <math>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}.</math>
 
# <math>x_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order.
 
# <math>x_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order.
 +
 +
We construct the following table for the case <math>x_1,</math> with all values in modulo <math>3:</math>
 +
<cmath>\begin{array}{c|c|c|c|c|c}
 +
\boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\
 +
\hline
 +
& & & & & \\ [-2ex]
 +
1 & 1 & 2 & 2 & 0 & \checkmark \\
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1 & 2 & 1 & 2 & 0 & \checkmark \\
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1 & 2 & 2 & 1 & 2 & \\
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2 & 1 & 1 & 2 & 1 & \\
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2 & 1 & 2 & 1 & 0 & \checkmark \\
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2 & 2 & 1 & 1 & 0 & \checkmark
 +
\end{array}</cmath>
  
 
<b>I am on my way. No edit please. A million thanks.</b>
 
<b>I am on my way. No edit please. A million thanks.</b>

Revision as of 01:49, 23 March 2021

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 \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\]

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

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 $x_1,$ with all values in modulo $3:$ \[\begin{array}{c|c|c|c|c|c} \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ \hline & & & & & \\ [-2ex] 1 & 1 & 2 & 2 & 0 & \checkmark \\ 1 & 2 & 1 & 2 & 0 & \checkmark \\ 1 & 2 & 2 & 1 & 2 & \\ 2 & 1 & 1 & 2 & 1 & \\ 2 & 1 & 2 & 1 & 0 & \checkmark \\ 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\]

I am on my way. No edit please. A million thanks.

~MRENTHUSIASM

Solution 3

See also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 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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