Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Video solution)
m (→‎Solution 5 (Factoring))
 
(22 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
Find the number of permutations <math>x_1, x_2, x_3, x_4, x_5</math> of numbers <math>1, 2, 3, 4, 5</math> such that the sum of five products <cmath>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</cmath>
 
Find the number of permutations <math>x_1, x_2, x_3, x_4, x_5</math> of numbers <math>1, 2, 3, 4, 5</math> such that the sum of five products <cmath>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2</cmath>
 +
is divisible by <math>3</math>.
  
 
==Solution 1==
 
==Solution 1==
−
Since <math>3</math> is one of the numbers, a product with a <math>3</math> in it is automatically divisible by <math>3</math>, so WLOG <math>x_3=3</math>, we will multiply by <math>5</math> afterward since any of <math>x_1, x_2, ..., x_5</math> would be <math>3</math>, after some cancelation we see that now all we need to find is the number of ways that <math>x_5x_1(x_4+x_2)</math> is divisible by <math>3</math>, since <math>x_5x_1</math> is never divisible by <math>3</math>, now we just need to find the number of ways <math>x_4+x_2</math> is divisible by <math>3</math>, after some calculation you will see that there are <math>16</math> ways to choose <math>x_1, x_2, x_4,</math> and <math>x_5</math> in this way. So the desired answer is <math>16 \times 5=\boxed{080}</math>.
+
Since <math>3</math> is one of the numbers, a product with a <math>3</math> in it is automatically divisible by <math>3,</math> so WLOG <math>x_3=3,</math> we will multiply by <math>5</math> afterward since any of <math>x_1, x_2, \ldots, x_5</math> would be <math>3,</math> after some cancelation we see that now all we need to find is the number of ways that <math>x_5x_1(x_4+x_2)</math> is divisible by <math>3,</math> since <math>x_5x_1</math> is never divisible by <math>3,</math> now we just need to find the number of ways <math>x_4+x_2</math> is divisible by <math>3.</math> Note that <math>x_2</math> and <math>x_4</math> can be <math>(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),</math> or <math>(5, 4).</math> We have <math>2</math> ways to designate <math>x_1</math> and <math>x_5</math> for a total of <math>8 \cdot 2 = 16.</math> So the desired answer is <math>16 \cdot 5=\boxed{080}.</math>
  
−
~ math31415926535
+
~math31415926535
  
−
==Solution 2==
+
~MathFun1000 (Rephrasing for clarity)
−
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_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order.
 
  
 +
==Solution 2 (Cyclic Symmetry and Casework)==
 +
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:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><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></li><p>
 +
  <li><math>x_2,x_3,x_4,x_5</math> are congruent to <math>1,2,1,2\pmod{3}</math> in some order.</li><p>
 +
</ol>
 
We construct the following table for the case <math>x_1=3,</math> with all values in modulo <math>3:</math>
 
We construct the following table for the case <math>x_1=3,</math> with all values in modulo <math>3:</math>
 
<cmath>\begin{array}{c||c|c|c|c|c||c}
 
<cmath>\begin{array}{c||c|c|c|c|c||c}
 
& & & & & & \\ [-2.5ex]
 
& & & & & & \\ [-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?} \\
+
\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
 
\hline
 
& & & & & & \\ [-2ex]
 
& & & & & & \\ [-2ex]
Line 25: Line 29:
 
6 & 2 & 2 & 1 & 1 & 0 & \checkmark
 
6 & 2 & 2 & 1 & 1 & 0 & \checkmark
 
\end{array}</cmath>
 
\end{array}</cmath>
−
For Row 1, <math>(x_2,x_3)</math> can be either <math>(1,4)</math> or <math>(4,1),</math> and <math>(x_4,x_5)</math> can be either <math>(2,5)</math> or <math>(5,2).</math> By the Multiplication Principle, Row 1 produces <math>2\cdot2=4</math> permutations. Similarly, Rows 2, 5, and 6 each produces <math>4</math> permutations.
+
For Row 1, <math>(x_2,x_3)</math> can be either <math>(1,4)</math> or <math>(4,1),</math> and <math>(x_4,x_5)</math> can be either <math>(2,5)</math> or <math>(5,2).</math> By the Multiplication Principle, Row 1 produces <math>2\cdot2=4</math> permutations. Similarly, Rows 2, 5, and 6 each produce <math>4</math> permutations.
  
−
Together, we get <math>4\cdot4=16</math> permutations for the case <math>x_1=3.</math> By the cyclic symmetry, the cases <math>x_2=3, x_3=3, x_4=3,</math> and <math>x_5=3</math> all have the same count, thus the total number of permutations <math>x_1, x_2, x_3, x_4, x_5</math> is <math>16\cdot5=\boxed{080}.</math>
+
Together, we get <math>4\cdot4=16</math> permutations for the case <math>x_1=3.</math> By the cyclic symmetry, the cases <math>x_2=3, x_3=3, x_4=3,</math> and <math>x_5=3</math> all have the same count. Therefore, the total number of permutations <math>x_1, x_2, x_3, x_4, x_5</math> is <math>16\cdot5=\boxed{080}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 36: Line 40:
  
 
We are left with <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math>.
 
We are left with <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math>.
−
We need <math>x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 (mod 3)</math>
+
We need <math>x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 \pmod{3}</math>.
−
The only way is when They are <math>(+1,-1)</math> or <math>(-1, +1)</math> (mod 3)
+
The only way is when They are <math>(+1,-1)</math> or <math>(-1, +1) \pmod{3}</math>.
  
−
The numbers left with us are <math>1,2,4,5</math> which are <math>+1,-1,+1,-1</math> (mod <math>3</math>) respectively.
+
The numbers left with us are <math>1,2,4,5</math> which are <math>+1,-1,+1,-1\pmod{3}</math> respectively.
  
 
<math>+1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) <math>= +1 \cdot +1 \cdot +1</math> <math>\;\;\; OR \;\;\;+1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) = <math>-1 \cdot -1 \cdot +1</math>.
 
<math>+1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) <math>= +1 \cdot +1 \cdot +1</math> <math>\;\;\; OR \;\;\;+1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) = <math>-1 \cdot -1 \cdot +1</math>.
Line 45: Line 49:
 
<math>-1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) <math>= -1 \cdot -1 \cdot -1</math> <math>\;\;\; OR \;\;\;-1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) = <math>-1 \cdot +1 \cdot +1</math>
 
<math>-1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) <math>= -1 \cdot -1 \cdot -1</math> <math>\;\;\; OR \;\;\;-1</math> (of <math>x_{4}x_{5}x_{1}</math> or <math>x_{5}x_{1}x_{2}</math>) = <math>-1 \cdot +1 \cdot +1</math>
  
−
But, as we have just two <math>+1's</math> and two <math>-1's</math>,
+
But, as we have just two <math>+1's</math> and two <math>-1's</math>.
−
Hence, We will have to take <math>+1 = +1 \cdot -1 \cdot -1</math> and <math>-1 = -1 \cdot +1 \cdot +1</math>
+
Hence, We will have to take <math>+1 = +1 \cdot -1 \cdot -1</math> and <math>-1 = -1 \cdot +1 \cdot +1</math>.
−
Among these two, we have a <math>+1</math> and <math>-1</math> in common, i.e. <math>(x_{5}, x_{1}) = (+1, -1) or (-1, +1)</math> (because <math>x_{1}</math> and <math>x_{5}</math>
+
Among these two, we have a <math>+1</math> and <math>-1</math> in common, i.e. <math>(x_{5}, x_{1}) = (+1, -1) or (-1, +1)</math> (because <math>x_{1}</math> and <math>x_{5}</math>.
−
are common in <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math> )
+
are common in <math>x_{4}x_{5}x_{1}</math> and <math>x_{5}x_{1}x_{2}</math>).
  
 
So, <math>(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}</math> i.e. <math>8</math> values.
 
So, <math>(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}</math> i.e. <math>8</math> values.
  
−
For each value of <math>(x_{5}, x_{1})</math> we get <math>2</math> values for <math>(x_{2}, x_{4})</math>
+
For each value of <math>(x_{5}, x_{1})</math> we get <math>2</math> values for <math>(x_{2}, x_{4})</math>.
 
Hence, in total, we have <math>8 \times 2 = 16</math> ways.
 
Hence, in total, we have <math>8 \times 2 = 16</math> ways.
  
−
But any of the <math>x_{i} 's</math> can be <math>3</math>
+
But any of the <math>x_{i} 's</math> can be <math>3</math>.
−
So, <math>16 \times 5 = \boxed{080}</math>
+
So, <math>16 \times 5 = \boxed{080}</math>.
  
 
-Arnav Nigam
 
-Arnav Nigam
 +
 +
==Solution 4 (Proportion)==
 +
WLOG, let <math>x_{3} = 3</math>. Then:
 +
<cmath>x_{1}x_{2}x_{3} + x_{2}x_{3}x_{4} + x_{3}x_{4}x_{5} + x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} = 3 (x_1 x_2 + x_2 x_4 + x_4 x_5) + x_5 x_1 (x_2  + x_4).</cmath>
 +
The sum is divisible by <math>3</math> if and only if  <math>x_2  + x_4</math>  is divisible by <math>3</math>.
 +
The possible sums of  <math>x_2  + x_4</math> are <math>1 + 2, 1 + 4, 1 + 5, 2 + 4, 2 + 5, 4 + 5.</math>
 +
Two of them are not multiples of <math>3</math>, but four of them are multiples.
 +
 +
A total number of  permutations is <math>5! = 120.</math>
 +
 +
<math>\frac {2}{3}</math> of this number, that is, <math>80,</math> give sums that are multiples of <math>3.</math>
 +
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 5 (Factoring)==
 +
 +
This is my first time doing a solution (feel free to edit it)
 +
 +
We have <cmath>x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.</cmath>
 +
 +
We have <math>5</math> numbers. Considering any <math>x</math> as <math>3,</math> we see that we are left with two terms that are not always divisible by <math>3,</math> which means that already gives us 5 options.
 +
Let's now consider <math>x_1 =3:</math>We are left with <cmath>3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.</cmath>
 +
The two terms left over are
 +
<cmath>x_2x_3x_4 + x_3x_4x_5 \equiv 0 \pmod{3}</cmath>
 +
since we already have used <math>3</math> the remaining numbers are <math>1,2,4,5.</math>
 +
 +
We now factor
 +
<cmath>\begin{align*}
 +
(x_2 + x_5)(x_3x_4) &\equiv 0 \pmod{3} \\
 +
(x_2 + x_5) &\equiv 0 \pmod{3}
 +
\end{align*}</cmath>
 +
since <math>1,2,4,5</math> are all not factors of <math>3.</math>
 +
 +
Now using the number <math>1,2,4,5,</math> we take two to get a number divisible by <math>3</math> for <math>(x_2 + x_5):</math>
 +
<cmath>\begin{align*}
 +
1+5 &\equiv 0 \pmod{3}, \\
 +
4+2 &\equiv 0 \pmod{3}, \\
 +
4+5 &\equiv 0 \pmod{3}, \\
 +
1+2 &\equiv 0 \pmod{3}.
 +
\end{align*}</cmath>
 +
We have <math>4</math> possibilities from above. Since we can also have <math>5+1</math> or <math>2+4,</math> there are <math>4\cdot2=8</math> possibilities in all.
 +
 +
Now using <cmath>(x_2 + x_5)(x_3x_4) \equiv 0 \pmod{3},</cmath> we have <math>(x_3x_4),</math> which results in <math>8</math> more possibilities of <math>2</math> times more. So, we get <math>2\cdot2\cdot4=16.</math>
 +
 +
Remember that <math>3</math> can be any of <math>5</math> different variables. So, we multiply by <math>5</math> to get the answer <math>\boxed{080}.</math>
  
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=HikWWhQlkVw
 
https://www.youtube.com/watch?v=HikWWhQlkVw
  
−
==See also==
+
==See Also==
 
{{AIME box|year=2021|n=II|num-b=2|num-a=4}}
 
{{AIME box|year=2021|n=II|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:52, 11 March 2023

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 $x_1, x_2, \ldots, x_5$ 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 $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \cdot 2 = 16.$ So the desired answer is $16 \cdot 5=\boxed{080}.$

~math31415926535

~MathFun1000 (Rephrasing for clarity)

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 $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 \pmod{3}$. The only way is when They are $(+1,-1)$ or $(-1, +1) \pmod{3}$.

The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1\pmod{3}$ 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

Solution 4 (Proportion)

WLOG, let $x_{3} = 3$. Then: \[x_{1}x_{2}x_{3} + x_{2}x_{3}x_{4} + x_{3}x_{4}x_{5} + x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} = 3 (x_1 x_2 + x_2 x_4 + x_4 x_5) + x_5 x_1 (x_2 + x_4).\] The sum is divisible by $3$ if and only if $x_2 + x_4$ is divisible by $3$. The possible sums of $x_2 + x_4$ are $1 + 2, 1 + 4, 1 + 5, 2 + 4, 2 + 5, 4 + 5.$ Two of them are not multiples of $3$, but four of them are multiples.

A total number of permutations is $5! = 120.$

$\frac {2}{3}$ of this number, that is, $80,$ give sums that are multiples of $3.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 5 (Factoring)

This is my first time doing a solution (feel free to edit it)

We have \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.\]

We have $5$ numbers. Considering any $x$ as $3,$ we see that we are left with two terms that are not always divisible by $3,$ which means that already gives us 5 options. Let's now consider $x_1 =3:$We are left with \[3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.\] The two terms left over are \[x_2x_3x_4 + x_3x_4x_5 \equiv 0 \pmod{3}\] since we already have used $3$ the remaining numbers are $1,2,4,5.$

We now factor \begin{align*} (x_2 + x_5)(x_3x_4) &\equiv 0 \pmod{3} \\ (x_2 + x_5) &\equiv 0 \pmod{3} \end{align*} since $1,2,4,5$ are all not factors of $3.$

Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \begin{align*} 1+5 &\equiv 0 \pmod{3}, \\ 4+2 &\equiv 0 \pmod{3}, \\ 4+5 &\equiv 0 \pmod{3}, \\ 1+2 &\equiv 0 \pmod{3}. \end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\cdot2=8$ possibilities in all.

Now using \[(x_2 + x_5)(x_3x_4) \equiv 0 \pmod{3},\] we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\cdot2\cdot4=16.$

Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $\boxed{080}.$

Video Solution

https://www.youtube.com/watch?v=HikWWhQlkVw

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_3&oldid=190817"