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)
(Added in Solution 2.)
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>
  
−
==Solution==
+
==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, ..., 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>.
  
 
~ math31415926535
 
~ math31415926535
 +
 +
==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</math> are <math>1,2,4,5</math> in some order. In modulo <math>3,</math> they are <math>1,2,1,2</math> in some order.
 +
 +
<b>I am on my way. No edit please. A million thanks.</b>
 +
 +
~MRENTHUSIASM
  
 
==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}}

Revision as of 00:07, 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 $x_2,x_3,x_4,x_5$ are $1,2,4,5$ in some order. In modulo $3,$ they are $1,2,1,2$ in some order.

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

~MRENTHUSIASM

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=149903"