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 ,
, 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.
,
,
,
Case : For
,
,
,
Case : For
,
Case :
,
,
, no solution
Case :
,
,
, no solution
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.