Difference between revisions of "2022 AIME II Problems/Problem 14"
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− | Notice that we must have <math>a = 1</math>; otherwise <math>1</math> cent stamp cannot be represented. At least <math>b-1</math> numbers of <math>1</math> cent stamps are needed to represent the values less than <math>b</math>. Using at most <math>c-1</math> stamps of value <math>1</math> and <math>b</math>, it can have all the values from <math>1</math> to <math>c-1</math> cents. Plus <math>\lfloor \frac{999}{c} \rfloor</math> stamps of value <math>c</math>, every value up to <math>1000</math> can 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> can be represented in sub-collections, while minimizing the number of stamps. | + | Notice that we must have <math>a = 1</math>; otherwise <math>1</math> cent stamp cannot be represented. At least <math>b-1</math> numbers of <math>1</math> cent stamps are needed to represent the values less than <math>b</math>. Using at most <math>c-1</math> stamps of value <math>1</math> and <math>b</math>, it can have all the values from <math>1</math> to <math>c-1</math> cents. Plus <math>\lfloor \frac{999}{c} \rfloor</math> stamps of value <math>c</math>, every value up to <math>1000</math> can be represented. |
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+ | <math>\hrule</math> | ||
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+ | Correction: | ||
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+ | This should be <math>\lfloor \frac{1000}{c} \rfloor</math>. The current function breaks when <math>c \mid 1000</math> and <math>b \mid c</math>. Take <math>c = 200</math> and <math>b = 20</math>. Then, we have <math>\lfloor \frac{999}{200} \rfloor = 4</math> stamps of value 200, <math>\lfloor \frac{199}{20} \rfloor = 9</math> stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is <math>200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999</math>, just shy of the needed 1000. As for the rest of solution, proceed similarly, except use <math>1000</math> instead of <math>999</math>. | ||
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+ | Also, some explanation: <math>b-1</math> one cent stamps cover all residues module <math>b</math>. Having <math>\lfloor \frac{c-1}{b} \rfloor</math> stamps of value b covers all residue classes modulo <math>c</math>. Finally, we just need <math>\lfloor \frac{1000}{c} \rfloor</math> to cover everything up to 1000. | ||
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+ | In addition, note that this function sometimes may not always minimize the number of stays required. This is due to the fact that the stamps of value <math>b</math> and of value <math>1</math> have the capacity to cover all values up to <math>c-1</math>. Thus, in certain cases, not all <math>\lfloor \frac{1000}{c} \rfloor</math> stamps of value c may be necessary, because the stamps of value <math>b</math> and 1 can replace one <math>c</math>. | ||
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+ | <math>\hrule</math> | ||
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+ | 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> can be represented in sub-collections, while minimizing the number of stamps. | ||
So, <math>f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1</math>. | So, <math>f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1</math>. |
Revision as of 17:57, 9 February 2024
Contents
[hide]Problem
For positive integers , , and with , consider collections of postage stamps in denominations , , and 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 cents, let be the minimum number of stamps in such a collection. Find the sum of the three least values of such that for some choice of and .
Solution
Notice that we must have ; otherwise cent stamp cannot be represented. At least numbers of cent stamps are needed to represent the values less than . Using at most stamps of value and , it can have all the values from to cents. Plus stamps of value , every value up to can be represented.
$\hrule$ (Error compiling LaTeX. Unknown error_msg)
Correction:
This should be . The current function breaks when and . Take and . Then, we have stamps of value 200, stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is , just shy of the needed 1000. As for the rest of solution, proceed similarly, except use instead of .
Also, some explanation: one cent stamps cover all residues module . Having stamps of value b covers all residue classes modulo . Finally, we just need to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stays required. This is due to the fact that the stamps of value and of value have the capacity to cover all values up to . Thus, in certain cases, not all stamps of value c may be necessary, because the stamps of value and 1 can replace one .
$\hrule$ (Error compiling LaTeX. Unknown error_msg)
Therefore using stamps of value , stamps of value , and stamps of value , all values up to can be represented in sub-collections, while minimizing the number of stamps.
So, .
. We can get the answer by solving this equation.
, or
,
,
,
, or .
,
,
The least values of are , , .
~isabelchen ~edited by bobjoebilly
Video Solution
~MathProblemSolvingSkills.com
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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