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Difference between revisions of "2022 AIME II Problems/Problem 8"
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==Solution==
 
==Solution==
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1. For <math>n</math> to be uniquely determined, <math>n</math> AND <math>n + 1</math> both need to be a multiple of <math>4, 5,</math> or <math>6.</math> Since either <math>n</math> or <math>n + 1</math> is odd, we know that either <math>n</math> or <math>n + 1</math> has to be a multiple of <math>5.</math> We can state the following cases:
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1. <math>n</math> is a multiple of <math>4</math> and <math>n+1</math> is a multiple of <math>5</math>
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2. <math>n</math> is a multiple of <math>6</math> and <math>n+1</math> is a multiple of <math>5</math>
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3. <math>n</math> is a multiple of <math>5</math> and <math>n+1</math> is a multiple of <math>4</math>
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4. <math>n</math> is a multiple of <math>5</math> and <math>n+1</math> is a multiple of <math>6</math>
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Solving for each case, we see that there are <math>20</math> possibilities for cases 1 and 3 each, and <math>30</math> possibilities for cases 2 and 4 each. However, we overcounted the cases where
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1. <math>n</math> is a multiple of <math>24</math> and <math>n+1</math> is a multiple of <math>5</math>
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2. <math>n</math> is a multiple of <math>5</math> and <math>n+1</math> is a multiple of <math>24</math>
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Each case has <math>10</math> possibilities.
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Adding all the cases and correcting for overcounting, we get <math>20 + 30 + 20 + 30 - 10 - 10 = \boxed {080}.</math>
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~Lucasfunnyface
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Side note: solution does not explain how we found the 20 possibilities, 30, possibilities, etc. It would be great if somebody added that in.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2022|n=II|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:21, 18 February 2022

Problem

Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real number $x$.

Solution

1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:

1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$

2. $n$ is a multiple of $6$ and $n+1$ is a multiple of $5$

3. $n$ is a multiple of $5$ and $n+1$ is a multiple of $4$

4. $n$ is a multiple of $5$ and $n+1$ is a multiple of $6$

Solving for each case, we see that there are $20$ possibilities for cases 1 and 3 each, and $30$ possibilities for cases 2 and 4 each. However, we overcounted the cases where

1. $n$ is a multiple of $24$ and $n+1$ is a multiple of $5$

2. $n$ is a multiple of $5$ and $n+1$ is a multiple of $24$

Each case has $10$ possibilities.

Adding all the cases and correcting for overcounting, we get $20 + 30 + 20 + 30 - 10 - 10 = \boxed {080}.$

~Lucasfunnyface

Side note: solution does not explain how we found the 20 possibilities, 30, possibilities, etc. It would be great if somebody added that in.

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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. AMC logo.png

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