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Difference between revisions of "2012 AIME I Problems/Problem 7"

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(Additional solution added)
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</asy></center>
 
</asy></center>
  
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== Solution ==
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== Solutions ==
 +
 
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===Solution 1===
 
Say the student in the center starts out with <math>a</math> coins, the students neighboring the center student each start with <math>b</math> coins, and all other students start out with <math>c</math> coins. Then the <math>a</math>-coin student has five neighbors, all the <math>b</math>-coin students have three neighbors, and all the <math>c</math>-coin students have four neighbors.
 
Say the student in the center starts out with <math>a</math> coins, the students neighboring the center student each start with <math>b</math> coins, and all other students start out with <math>c</math> coins. Then the <math>a</math>-coin student has five neighbors, all the <math>b</math>-coin students have three neighbors, and all the <math>c</math>-coin students have four neighbors.
  
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</cmath>
 
</cmath>
  
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Solving these equations, we see that <math>\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.</math> Also, the total number of coins is <math>a + 5b + 10c = 3360,</math> so <math>a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280.}</math>
+
Solving these equations, we see that <math>\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.</math> Also, the total number of coins is <math>a + 5b + 10c = 3360,</math> so <math>a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.</math>
 +
 
 +
===Solution 2===
 +
Since the students give the same number of gifts of coins as they receive and still end up the same number of coins, we can assume that every gift of coins has the same number of coins. Let <math>x</math> be the number of coins in each gift of coins. There <math>10</math> people who give <math>4</math> gifts of coins, <math>5</math> people who give <math>3</math> gifts of coins, and <math>1</math> person who gives <math>5</math> gifts of coins. Thus,
 +
 
 +
<cmath>
 +
\begin{align*}
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10(4x)+5(3x)+5x &= 3360\\
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40x+15x+5x &= 3360\\
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60x &= 3360\\
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x &= 56
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\end{align*}
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</cmath>
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Therefore the answer is <math>5(56) = \boxed{280}.</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2012|n=I|num-b=6|num-a=8}}

Revision as of 18:04, 14 March 2013

Problem 7

At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

[asy] import cse5; unitsize(6mm); defaultpen(linewidth(.8pt)); dotfactor = 8; pathpen=black; pair A = (0,0); pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342); pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306); pair M = 6.4*dir(54), N = 6.4*dir(126), O = 6.4*dir(198), P = 6.4*dir(270), L = 6.4*dir(342); pair[] dotted = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P}; D(A--B--H--M); D(A--C--H--N); D(A--F--G--L); D(A--E--K--P); D(A--D--J--O); D(B--G--M); D(F--K--L); D(E--J--P); D(O--I--D); D(C--I--N); D(L--M--N--O--P--L); dot(dotted); [/asy]

Solutions

Solution 1

Say the student in the center starts out with $a$ coins, the students neighboring the center student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$-coin student has five neighbors, all the $b$-coin students have three neighbors, and all the $c$-coin students have four neighbors.

Now in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must be equal to the number received, and thus

\begin{align*} a &= 5 \cdot \frac{b}{3}\\ b &= \frac{a}{5} + 2 \cdot \frac{c}{4}\\ c &= 2 \cdot \frac{c}{4} + 2 \cdot \frac{b}{3}. \end{align*}

Solving these equations, we see that $\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.$ Also, the total number of coins is $a + 5b + 10c = 3360,$ so $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.$

Solution 2

Since the students give the same number of gifts of coins as they receive and still end up the same number of coins, we can assume that every gift of coins has the same number of coins. Let $x$ be the number of coins in each gift of coins. There $10$ people who give $4$ gifts of coins, $5$ people who give $3$ gifts of coins, and $1$ person who gives $5$ gifts of coins. Thus,

\begin{align*} 10(4x)+5(3x)+5x &= 3360\\ 40x+15x+5x &= 3360\\ 60x &= 3360\\ x &= 56 \end{align*}

Therefore the answer is $5(56) = \boxed{280}.$

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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