2022 AIME II Problems/Problem 8

Revision as of 21:21, 18 February 2022 by Lucasfunnyface (talk | contribs) (Solution)

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 (ProblemsAnswer KeyResources)
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

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