Difference between revisions of "2022 AIME II Problems/Problem 14"
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<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> | ||
− | The <math>3</math> least values of <math>c</math> | + | The <math>3</math> least values of <math>c</math> are <math>11</math>, <math>88</math>, <math>89</math>. <math>11 + 88+ 89 = \boxed{\textbf{188}}</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 02:19, 13 June 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
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
,
,
.
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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