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Difference between revisions of "2011 AIME I Problems/Problem 5"

(Modulus theory shows that the answer is 144.)
 
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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 mod 3. We see that the numbers 1, 4, and 7 are congruent to 1 (mod 3), that the numbers 2, 5, and 8 are congruent to 2 (mod 3), and that the numbers 3, 6, and 9 are congruent to 0 (mod 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 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.  
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== Problem ==
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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.  
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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.
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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:
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The first digit is predetermined because we want to avoid strings that rotate to become indistinguishable, so we have one option as a choice for the first digit.
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The second and third digits can each be any of the three digits congruent to the correct mod 3 number according to the trio arrangement of this nonagon.
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== Solution 1 ==
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The fourth, fifth, and sixth digits can each be chosen from either of the two remaining options following the mod 3 trio arrangement.  
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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>.
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The last three digits will be determined by the first six and we have no remaining choices left for them.
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Thus we have 2 (trio arrangements) * 1 (choice for the first digit) * 3 (choices for the second digit) * 3 (choices for the third digit) * 2 (choices for the fourth digit) * 2 (choices for the fifth digit) * 2 (choices for the sixth digit) * 1 (choice for the seventh digit) * 1 (choice for the eighth digit) * 1 (choice for the ninth digit) = 3^2 * 2^4 = '''144''' possibilities
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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.  
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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.  
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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>.
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== Solution 2==
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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.
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==Video solution==
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https://www.youtube.com/watch?v=vkniYGN45F4
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== See also ==
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{{AIME box|year=2011|n=I|num-b=4|num-a=6}}
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{{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)
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