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2005 AMC 8 Problems/Problem 20: Difference between revisions

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==Solution==
==Solution==
Alice moves <math>5k</math> steps and Bob moves <math>9k</math> steps, where <math>k</math> is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, <math>14k</math>, is a multiple of <math>12</math>. Since <math>14</math> has a factor <math>2</math>, <math>k</math> must have a factor of <math>6</math>. The smallest number of turns that is a multiple of <math>6</math> is <math>\boxed{\textbf{(A)}\ 6}</math>.
Alice moves <math>5k</math> steps and Bob moves <math>9k</math> steps, where <math>k</math> is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, <math>14k</math>, is a multiple of <math>12</math>. Since this number must be a multiple of <math>12</math>, as stated in the previous sentence, <math>14</math> has a factor <math>2</math>, <math>k</math> must have a factor of <math>6</math>. The smallest number of turns that is a multiple of <math>6</math> is <math>\boxed{\textbf{(A)}\ 6}</math>.
 
Another solution (Modular Arithmetic) Step 1: Write their positions after
𝑘
k turns.
 
Alice: moves 5 steps clockwise each turn.
Position after
𝑘
k turns:
 
𝐴
(
𝑘
)
=
12
+
5
𝑘
(
m
o
d
12
)
A(k)=12+5k(mod12)
 
Bob: moves 9 steps counterclockwise each turn.
Moving 9 steps counterclockwise is the same as moving +3 clockwise (since
12
−
9
=
3
12−9=3).
Position after
𝑘
k turns:
 
𝐵
(
𝑘
)
=
12
+
3
𝑘
(
m
o
d
12
)
B(k)=12+3k(mod12)
Step 2: Find when they meet.
 
They meet when:
 
𝐴
(
𝑘
)
≡
𝐵
(
𝑘
)
(
m
o
d
12
)
A(k)≡B(k)(mod12)
 
So:
 
12
+
5
𝑘
≡
12
+
3
𝑘
(
m
o
d
12
)
12+5k≡12+3k(mod12)
 
Cancel the 12’s:
 
5
𝑘
≡
3
𝑘
(
m
o
d
12
)
5k≡3k(mod12)
2
𝑘
≡
0
(
m
o
d
12
)
2k≡0(mod12)
Step 3: Solve for
𝑘
k.
 
That means
2
𝑘
2k is divisible by 12:
 
2
𝑘
≡
0
(
m
o
d
12
)
⇒
𝑘
≡
0
(
m
o
d
6
)
2k≡0(mod12)⇒k≡0(mod6)
 
So the smallest positive
𝑘
k is:
 
𝑘
=
6
k=6
 
✅ They meet after 6 turns → Answer: (A) 6  - timi821


==See Also==
==See Also==
{{AMC8 box|year=2005|num-b=19|num-a=21}}
{{AMC8 box|year=2005|num-b=19|num-a=21}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 15:10, 3 September 2025

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

Another solution (Modular Arithmetic) Step 1: Write their positions after 𝑘 k turns.

Alice: moves 5 steps clockwise each turn. Position after 𝑘 k turns:

𝐴 ( 𝑘 ) = 12 + 5 𝑘 ( m o d 12 ) A(k)=12+5k(mod12)

Bob: moves 9 steps counterclockwise each turn. Moving 9 steps counterclockwise is the same as moving +3 clockwise (since 12 − 9 = 3 12−9=3). Position after 𝑘 k turns:

𝐵 ( 𝑘 ) = 12 + 3 𝑘 ( m o d 12 ) B(k)=12+3k(mod12) Step 2: Find when they meet.

They meet when:

𝐴 ( 𝑘 ) ≡ 𝐵 ( 𝑘 ) ( m o d 12 ) A(k)≡B(k)(mod12)

So:

12 + 5 𝑘 ≡ 12 + 3 𝑘 ( m o d 12 ) 12+5k≡12+3k(mod12)

Cancel the 12’s:

5 𝑘 ≡ 3 𝑘 ( m o d 12 ) 5k≡3k(mod12) 2 𝑘 ≡ 0 ( m o d 12 ) 2k≡0(mod12) Step 3: Solve for 𝑘 k.

That means 2 𝑘 2k is divisible by 12:

2 𝑘 ≡ 0 ( m o d 12 ) ⇒ 𝑘 ≡ 0 ( m o d 6 ) 2k≡0(mod12)⇒k≡0(mod6)

So the smallest positive 𝑘 k is:

𝑘 = 6 k=6

✅ They meet after 6 turns → Answer: (A) 6 - timi821

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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