2012 AIME I Problems/Problem 7
Problem 7
At each of the sixteen circles in the network below stands a student. A total of 
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]](http://latex.artofproblemsolving.com/3/3/a/33abde80848746d240bfc392b8997b80cf3bccd5.png)
Solutions
Solution 1
Say the student in the center starts out with 
coins, the students neighboring the center student each start with 
coins, and all other students start out with 
coins. Then the -coin student has five neighbors, all the -coin students have three neighbors, and all the -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
Solving these equations, we see that 
Also, the total number of coins is 
so 
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 
be the number of coins in each gift of coins. There 
people who give 
gifts of coins, 
people who give 
gifts of coins, and 
person who gives gifts of coins. Thus,
Therefore the answer is 
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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