Difference between revisions of "2022 AIME II Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
| − | Notice that <math>a | + | Notice that we must have <math>a = 1</math>, or else <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 is able to 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> 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. |
| − | < | + | |
| + | So, <math>f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1</math> | ||
| + | |||
| + | <math>\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97</math> | ||
| + | |||
| + | We can get the answer by solving this equation. | ||
To be continued...... | To be continued...... | ||
Revision as of 12:20, 19 February 2022
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 1
Notice that we must have 
, or else 
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 is able to have all the values from to cents. Plus 
stamps of value , every value up to is able to be represented. Therefore using stamps of value , 
stamps of value , and stamps of value all values up to are able to be represented in sub-collections, while minimizing the number of stamps.
So, 
We can get the answer by solving this equation.
To be continued......
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
