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 "2011 AIME I Problems/Problem 5"

(→‎Solution: LaTeX and cleanup in general)
(→‎Quick Solution)
 
(19 intermediate revisions by 7 users not shown)
Line 2: Line 2:
 
The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3.  Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane.  Find the number of distinguishable acceptable arrangements.
 
The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3.  Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane.  Find the number of distinguishable acceptable arrangements.
  
−
== Solution ==
+
 
 +
== Solution 1 ==
 
First, we determine which possible combinations of digits <math>1</math> through <math>9</math> will yield sums that are multiples of <math>3</math>. It is simplest to do this by looking at each of the digits <math>\bmod{3}</math>.
 
First, we determine which possible combinations of digits <math>1</math> through <math>9</math> will yield sums that are multiples of <math>3</math>. It is simplest to do this by looking at each of the digits <math>\bmod{3}</math>.
  
−
We see that the numbers <math>1, 4,</math> and <math>7</math> are congruent to <math>1 \pmod{3}</math>, that the numbers <math>2, 5,</math> and <math>8</math> are congruent to <math>2 \pmod{3}</math>, and that the numbers <math>3, 6,</math> and <math>9</math> are congruent to <math>0 \pmod{3}</math>. In order for a sum of three of these numbers to be a multiple of three, the mod <math>3</math> sum must be congruent to <math>0</math>. Quick inspection reveals that the only possible combinations are <math>0+0+0, 1+1+1, 2+2+2,</math> and <math>0+1+2</math>. However, every set of three consecutive vertices must sum to a multiple of three, so using any of <math>0+0+0, 1+1+1</math>, or <math>2+2+2</math> would cause an adjacent sum to include exactly 2 digits with the same mod 3 value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different mod 3 values.  
+
We see that the numbers <math>1, 4,</math> and <math>7</math> are congruent to <math>1 \pmod{3}</math>, that the numbers <math>2, 5,</math> and <math>8</math> are congruent to <math>2 \pmod{3}</math>, and that the numbers <math>3, 6,</math> and <math>9</math> are congruent to <math>0 \pmod{3}</math>. In order for a sum of three of these numbers to be a multiple of three, the mod <math>3</math> sum must be congruent to <math>0</math>. Quick inspection reveals that the only possible combinations are <math>0+0+0, 1+1+1, 2+2+2,</math> and <math>0+1+2</math>. However, every set of three consecutive vertices must sum to a multiple of three, so using any of <math>0+0+0, 1+1+1</math>, or <math>2+2+2</math> would cause an adjacent sum to include exactly <math>2</math> digits with the same <math>\bmod{3}</math> value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different <math> \bmod{3} </math> values.
 +
 
 +
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to <math> 1 \pmod{3} </math> can be located counterclockwise of a digit congruent to <math> 0 </math> and clockwise of a digit congruent to <math> 2 \pmod{3} </math>, or the reverse can be true.
 +
 
 +
We set the first digit as <math> 3 </math> avoid overcounting rotations, so we have one option as a choice for the first digit.  The other two <math> 0 \pmod{3} </math> numbers can be arranged in <math> 2!=2 </math> ways.  The three <math> 1 \pmod{3}</math> and three <math> 2 \pmod{3} </math> can both be arranged in <math>3!=6</math> ways.  Therefore, the desired result is <math> 2(2 \times 6 \times 6)=\boxed{144} </math>.
  
−
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to 1 (mod 3) can be located counterclockwise of a digit congruent to 0 and clockwise of a digit congruent to 2, or the reverse can be true.  
+
== Solution 2==
 +
Notice that there are three triplets of congruent integers <math>\mod 3</math>: <math>(1,4,7),</math> <math>(2,5,8),</math> and <math>(3,6,9).</math> There are <math>3!</math> ways to order each of the triplets individually and <math>3!</math> ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are <math>(3!)^4/9=\boxed{144}</math> total arrangements.
  
−
The nonagon can be rotated, so if we find all possible strings beginning with one particular digit, we have found all indistinguishable arrangements. For each of the two trio arrangements, we find 72 possible strings:
+
==Video solution==
  
−
The first digit is predetermined as 3 because we want to avoid strings that rotate to become indistinguishable, so we have one option as a choice for the first digit. The other two 0 (mod 3) numbers can be arranged in 2!=2 ways.  The three 1 (mod 3) and three  2 (mod 3) can both be arranged in 3!=6 ways. Therefore, the desired result is 2(2*6*6)=144.
+
https://www.youtube.com/watch?v=vkniYGN45F4
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2011|n=I|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Latest revision as of 04:29, 17 December 2023

Problem

The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements.


Solution 1

First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$. It is simplest to do this by looking at each of the digits $\bmod{3}$.

We see that the numbers $1, 4,$ and $7$ are congruent to $1 \pmod{3}$, that the numbers $2, 5,$ and $8$ are congruent to $2 \pmod{3}$, and that the numbers $3, 6,$ and $9$ are congruent to $0 \pmod{3}$. In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$. Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$. However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$, or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\bmod{3}$ values.

We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \pmod{3}$, or the reverse can be true.

We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit. The other two $0 \pmod{3}$ numbers can be arranged in $2!=2$ ways. The three $1 \pmod{3}$ and three $2 \pmod{3}$ can both be arranged in $3!=6$ ways. Therefore, the desired result is $2(2 \times 6 \times 6)=\boxed{144}$.

Solution 2

Notice that there are three triplets of congruent integers $\mod 3$: $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\boxed{144}$ total arrangements.

Video solution

https://www.youtube.com/watch?v=vkniYGN45F4

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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=2011_AIME_I_Problems/Problem_5&oldid=207754"