Difference between revisions of "2022 AIME II Problems/Problem 14"

Revision as of 11:53, 19 February 2022

Problem

For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ 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 $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ .

Solution 1

Notice that we must have $a = 1$ , or else $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$ . Using at most $c-1$ stamps of value $1$ and $b$ , it is able to have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ , every value up to $1000$ is able to be represented. Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ , $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ all values up to $1000$ are able to be represented in sub-collections, while minimizing the number of stamps.

So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

We can get the answer by solving this equation.

$c > \lfloor \frac{c-1}{b} \rfloor + b-1$

$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$

$c^2 - 97c + 999 > 0$ , $c > 85.3$ , $c < 11.7$

$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$

$97 > \frac{999}{c}$ , $c>10.3$

Case $1$ : For $10.3 < c < 11.7$ , $c = 11$ , $\lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97$

$\lfloor \frac{10}{b} \rfloor + b = 8$ , $b=7$

Case $2$ : For $c>85.3$ ,

Case $2.1$ : $c = 86$ , $\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97$

$\lfloor \frac{85}{b} \rfloor + b = 87$ , $b=87 > c$ , no solution

Case $2.2$ : $c = 87$ , $\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97$

$\lfloor \frac{86}{b} \rfloor + b = 87$ , $b=86$ or $1$ , solution is $b = 86$

Case $2.3$ : $c = 88$ , $\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97$

$\lfloor \frac{87}{b} \rfloor + b = 87$ , $b=86$

To be continued......

~isabelchen

@@ Line 35: / Line 35: @@
 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=87 > c</math>, no solution
+<math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=86</math> or <math>1</math>, solution is <math>b = 86</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>
+<math>\lfloor \frac{87}{b} \rfloor + b = 87</math>, <math>b=86</math>
 To be continued......

2022 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AIME II Problems/Problem 14"

Revision as of 11:53, 19 February 2022

Problem

Solution 1

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