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

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m (→‎Solution)
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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>\text{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>\text{Case } 1:</math> <math>10.3 < c < 11.7 \rightarrow c = 11 \rightarrow \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>
  
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  <math>c = 86 \rightarrow</math> <math>\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97</math>
 
  <math>c = 86 \rightarrow</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 \rightarrow \text{no solution}</math>
  
 
  <math>c = 87 \rightarrow</math> <math>\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97</math>
 
  <math>c = 87 \rightarrow</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>. We cannot have <math>b=1</math> since it doesn't satisfy <math>a<b</math>, and note that if  
+
  <math>\lfloor \frac{86}{b} \rfloor + b = 87</math>, <math>b=86</math> or <math>1</math>. <math>\text{We cannot have }b=1 \text{ since it doesn't satisfy } a<b \text{, and note that if } b=86 \text{ we can have 10 coins of value } c, 1 \text{ of } b, \text{ and } 85</math> <math>\text{of } a \text{ for a total of 96 coins and still be able to make every value from 1 to 1000. Thus } c=87 \text{ yields no solution.}</math>
−
<math>b=86</math> we can have <math>10</math> coins of value <math>c</math>, <math>1</math> of <math>b</math>, and <math>85</math> of <math>a</math> for a total of <math>96</math> coins and still be able to make  
 
−
every value from <math>1</math> to <math>1000</math>. Thus <math>c=87</math> yields no solution.
 
  
 
  <math>c = 88 \rightarrow</math> <math>\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97</math>
 
  <math>c = 88 \rightarrow</math> <math>\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97</math>

Revision as of 16:29, 3 July 2023

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

Notice that we must have $a = 1$, otherwise $1$ 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 can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, every value up to $1000$ can 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 $1000$ can 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.

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

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

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

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

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

$\text{Case } 1:$ $10.3 < c < 11.7 \rightarrow c = 11 \rightarrow \lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97$
$\lfloor \frac{10}{b} \rfloor + b = 8$, $b=7$

$\text{Case } 2: c>85.3$

$c = 86 \rightarrow$ $\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97$
$\lfloor \frac{85}{b} \rfloor + b = 87$, $b=87 > c \rightarrow \text{no solution}$

$c = 87 \rightarrow$ $\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97$
$\lfloor \frac{86}{b} \rfloor + b = 87$, $b=86$ or $1$. $\text{We cannot have }b=1 \text{ since it doesn't satisfy } a<b \text{, and note that if } b=86 \text{ we can have 10 coins of value } c, 1 \text{ of } b, \text{ and } 85$ $\text{of } a \text{ for a total of 96 coins and still be able to make every value from 1 to 1000. Thus } c=87 \text{ yields no solution.}$

$c = 88 \rightarrow$ $\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97$
$\lfloor \frac{87}{b} \rfloor + b = 87$, $b=86$

$c = 89 \rightarrow$ $\lfloor \frac{999}{89} \rfloor + \lfloor \frac{88}{b} \rfloor + b-1 = 97$
$\lfloor \frac{88}{b} \rfloor + b = 87$, $b=86$

The $3$ least values of $c$ are $11$, $88$, $89$. $11 + 88+ 89 = \boxed{\textbf{188}}$

~isabelchen ~edited by bobjoebilly

Video Solution

https://youtu.be/jptMMVCuj34

~MathProblemSolvingSkills.com


See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AIME Problems and Solutions

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