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Difference between revisions of "2022 AIME II Problems/Problem 14"

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<math>\lfloor \frac{85}{b} \rfloor + b = 87</math>, <math>b=87 > c</math>, no solution
 
<math>\lfloor \frac{85}{b} \rfloor + b = 87</math>, <math>b=87 > c</math>, no solution
 +
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Case <math>2.2</math>: <math>c = 87</math>, <math>\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97</math>
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<math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=87 > c</math>, no solution
  
 
To be continued......
 
To be continued......

Revision as of 12:50, 19 February 2022

Problem

For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$.

Solution 1

Notice that we must have $a = 1$, or else $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it is able to have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, every value up to $1000$ is able to be represented. Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$, and $b-1$ stamps of value $1$ all values up to $1000$ are able to be represented in sub-collections, while minimizing the number of stamps.

So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

We can get the answer by solving this equation.

$c > \lfloor \frac{c-1}{b} \rfloor + b-1$

$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

$c^2 - 97c + 999 > 0$, $c > 85.3$, $c < 11.7$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$

$97 > \frac{999}{c}$, $c>10.3$

Case $1$: For $10.3 < c < 11.7$, $c = 11$, $\lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97$

$\lfloor \frac{10}{b} \rfloor + b = 8$, $b=7$

Case $2$: For $c>85.3$,

Case $2.1$: $c = 86$, $\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97$

$\lfloor \frac{85}{b} \rfloor + b = 87$, $b=87 > c$, no solution

Case $2.2$: $c = 87$, $\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97$

$\lfloor \frac{86}{b} \rfloor + b = 87$, $b=87 > c$, no solution

To be continued......

~isabelchen

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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