Difference between revisions of "2022 AIME II Problems/Problem 14"
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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. | 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> | + | So, <math>f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1</math>, <math>b<c-1</math> |
<math>\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97</math> | <math>\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97</math> | ||
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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> | Case <math>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>, | + | <math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=86</math> or <math>1</math>, <math>c-1=86</math>, <math>a=1</math>, no solution |
Case <math>2.3</math>: <math>c = 88</math>, <math>\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97</math> | Case <math>2.3</math>: <math>c = 88</math>, <math>\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97</math> | ||
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<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> | ||
− | + | Case <math>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> | ||
+ | |||
+ | The <math>3</math> least values of <math>c</math> is <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 12:59, 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 :
,
,
or
,
,
, no solution
Case :
,
,
Case :
,
,
The least values of
is
,
,
.
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.