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"

(→‎Solution 3)
(→‎Solution 3)
Line 39: Line 39:
 
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</math> (mod <math>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> or <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>+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>   or   <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>-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> or <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>-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>   or   <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>
  
 
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>,

Revision as of 01:04, 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

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 $8 \times 2 = \boxed{16}$ ways.

-Arnav Nigam

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=149930"