Maths Olympiad Prep

Library / /720 of 740

, 2018

Combinatorics Difficulty 5.9 AIME, harder Prove it United States

Problem:

Let nn be a positive integer. Let there be PnP_{n} ways for Pretty Penny to make exactly nn dollars out of quarters, dimes, nickels, and pennies. Also, let there be BnB_{n} ways for Beautiful Bill to make exactly nn dollars out of one dollar bills, quarters, dimes, and nickels. As nn goes to infinity, the sequence of fractions PnBn\frac{P_{n}}{B_{n}} approaches a real number cc. Find cc.

Note: Assume both Pretty Penny and Beautiful Bill each have an unlimited number of each type of coin. Pennies, nickels, dimes, quarters, and dollar bills are worth 1,5,10,25,1001,5,10,25,100 cents respectively.

Proposed by: James Lin

Solution

Solution:

Let dxd_{x} be the number ways to make exactly xx cents using only dimes and nickels. It is easy to see that when xx is a multiple of 55,
dx=x10+1 d_{x} = \left\lfloor \frac{x}{10} \right\rfloor + 1
Now, let cxc_{x} be the number of ways to make exactly xx cents using only quarters, dimes and nickels. Again, it is easy to see that when xx is a multiple of 55,
cx=cx25+dx c_{x} = c_{x-25} + d_{x}
(We can either use 1 or more quarters, which corresponds to the cx25c_{x-25} term, or we can use 0 quarters, which corresponds to the dxd_{x} term.) Combining these two equations, we see that cxc_{x} can be approximated by a polynomial of degree 2. (In fact, we get five different approximations of cxc_{x}, depending on the value of xmod25x \bmod 25, but they all only differ by a constant, which will not affect the limit case.) We also see that
Bn=c100n+c100(n1)++c0 B_{n} = c_{100 n} + c_{100(n-1)} + \ldots + c_{0}
and
Pn=c100n+c100n5++c0 P_{n} = c_{100 n} + c_{100 n - 5} + \ldots + c_{0}
Suppose aa is the value such that limncnan2=1\lim_{n \rightarrow \infty} \frac{c_{n}}{a n^{2}} = 1. Then
limnBnPn=limnk=0n/100a(100k)2k=0n/5a(5k)2=limn400n100(n100+1)(2n100+1)n5(n5+1)(2n5+1)=20. \lim_{n \rightarrow \infty} \frac{B_{n}}{P_{n}} = \lim_{n \rightarrow \infty} \frac{\sum_{k=0}^{\lfloor n / 100 \rfloor} a (100k)^{2}}{\sum_{k=0}^{\lfloor n / 5 \rfloor} a (5k)^{2}} = \lim_{n \rightarrow \infty} \frac{400 \cdot \frac{n}{100} \left( \frac{n}{100} + 1 \right) \left( 2 \cdot \frac{n}{100} + 1 \right)}{\frac{n}{5} \left( \frac{n}{5} + 1 \right) \left( 2 \cdot \frac{n}{5} + 1 \right)} = 20.

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