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

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==Solution 1==
 
==Solution 1==
  
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Notice that <math>a</math> must equal to <math>1</math>, or else the value <math>1</math> cent isn't able to be represented. At least <math>b-1</math> numbers of <math>1</math> cent stamps will be needed. Using at most <math>c-1</math> stamps of value <math>1</math> and <math>b</math>, it is able to have all the values from <math>1</math> to <math>c-1</math> cents.
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Notice that <math>a</math> must equal to <math>1</math>, or else the value <math>1</math> cent isn't able to be represented. At least <math>b-1</math> numbers of <math>1</math> cent stamps will be needed. Using at most <math>c-1</math> stamps of value <math>1</math> and <math>b</math>, it is able to have all the values from <math>1</math> to <math>c-1</math> cents. Adding in <math>\lfloor \frac{999}{c} \rfloor</math> stamps of value <math>c</math>, every value up to <math>1000</math> is able to be represented. Therefore using <math>\lfloor \frac{999}{c} \rfloor</math> stamps of value <math>c</math>, <math>\lfloor \frac{c-1}{b} \rfloor</math> stamps of value <math>b</math>, and <math>b-1</math> stamps of value <math>1</math> all values up to <math>1000</math> are able to be represented in sub-collections, while minimizing the number of stamps.
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<cmath>\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97</cmath>
  
 
To be continued......
 
To be continued......

Revision as of 12:10, 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 $a$ must equal to $1$, or else the value $1$ cent isn't able to be represented. At least $b-1$ numbers of $1$ cent stamps will be needed. 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. Adding in $\lfloor \frac{999}{c} \rfloor$ 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. \[\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97\]

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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