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

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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>
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*One way to make this more rigorous (answering the concern in Solution 3) is to let <math>a</math>, <math>b</math>, and <math>c</math> represent the <math>total</math> number of coins in the rings. Then, you don't care about individual people in the passing, you just know that each ring gets some coins and loses some coins, which must cancel each other out.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 12:10, 4 March 2018

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 $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.$

  • One way to make this more rigorous (answering the concern in Solution 3) is to let $a$, $b$, and $c$ represent the $total$ number of coins in the rings. Then, you don't care about individual people in the passing, you just know that each ring gets some coins and loses some coins, which must cancel each other out.

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 $5(56) = \boxed{280}.$

Solution 3

The assumption in Solution 1 (the students neighboring the center student each start with $b$ coins, why?) and Solution 2 (every gift of coins has the same number of coins, why?) seems not natural. This is a more strict solution without such assumption. Mark the number of coins from inside to outside as $a$, $b_1$, $b_2$, $b_3$, $b_4$, $b_5$, $c_1$, $c_2$, $c_3$, $c_4$, $c_5$, $d_1$, $d_2$, $d_3$, $d_4$, $d_5$. Then, we can get that \begin{align*} d_1 &= d_5/4 + d_2/4 + c_1/4 + c_2/4\\ d_2 &= d_1/4 + d_3/4 + c_2/4 + c_3/4\\ d_3 &= d_2/4 + d_4/4 + c_3/4 + c_4/4\\ d_4 &= d_3/4 + d_5/4 + c_4/4 + c_5/4\\ d_5 &= d_4/4 + d_1/4 + c_5/4 + c_1/4\\ \end{align*} Add them, let $D = d_1 + d_2 + d_3 + d_4 + d_5$, $C = c_1 + c_2 + c_3 + c_4 + c_5$, then $D = D/4 + D/4 + C/4 + C/4$, $D = C$. Repeat in the same way, we can get $C = D/4 + D/4 + B/3 + B/3$, $B = \frac{3D}{4}$, $B = C/4 + C/4 + a$, $a = \frac{D}{4}$. Then, with $a + B + C + D = 3360$, $D = 1120$, $a = \boxed{280}$.

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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