Difference between revisions of "2022 AIME II Problems/Problem 14"
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<math>97 > \frac{999}{c}</math>, <math>c>10.3</math> | <math>97 > \frac{999}{c}</math>, <math>c>10.3</math> | ||
− | <math> | + | <math>{Case 1:}</math> For <math>10.3 < c < 11.7</math>, <math>c = 11</math>, <math>\lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97</math> |
<math>\lfloor \frac{10}{b} \rfloor + b = 8</math>, <math>b=7</math> | <math>\lfloor \frac{10}{b} \rfloor + b = 8</math>, <math>b=7</math> | ||
− | <math> | + | <math>{Case 2:}</math> For <math>c>85.3</math>, |
− | <math> | + | <math>{Case 2.1:}</math> <math>c = 86</math>, <math>\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97</math> |
<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 | ||
− | <math> | + | <math>{Case 2.2:}</math> <math>c = 87</math>, <math>\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97</math> |
<math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=86</math> or <math>1</math>, neither values satisfy <math>a < b < c-1</math>, no solution | <math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=86</math> or <math>1</math>, neither values satisfy <math>a < b < c-1</math>, no solution | ||
− | <math> | + | <math>{Case 2.3:}</math> <math>c = 88</math>, <math>\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97</math> |
<math>\lfloor \frac{87}{b} \rfloor + b = 87</math>, <math>b=86</math> | <math>\lfloor \frac{87}{b} \rfloor + b = 87</math>, <math>b=86</math> | ||
− | <math> | + | <math>{Case 2.4:}</math> <math>c = 89</math>, <math>\lfloor \frac{999}{89} \rfloor + \lfloor \frac{88}{b} \rfloor + b-1 = 97</math> |
<math>\lfloor \frac{88}{b} \rfloor + b = 87</math>, <math>b=86</math> | <math>\lfloor \frac{88}{b} \rfloor + b = 87</math>, <math>b=86</math> | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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==Video Solution== | ==Video Solution== |
Revision as of 06:51, 2 February 2023
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. 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
,
For , , ,
For ,
, , , no solution
, , or , neither values satisfy , no solution
, ,
, ,
The least values of are , , .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.