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Difference between revisions of "2015 AIME II Problems/Problem 8"

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<cmath>\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.</cmath>
 
<cmath>\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.</cmath>
  
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<math></math>2a - 3<math>, </math>2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow<cmath>
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<cmath>2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow</cmath>
−
</cmath>(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).<cmath>
+
<cmath>(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).</cmath>
−
</cmath>(a, b) = (2, 2), (2, 3), (3, 2).<cmath>
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<cmath>(a, b) = (2, 2), (2, 3), (3, 2).</cmath>
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</cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.<cmath>
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<cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.</cmath>
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</cmath>\frac{31}{5} \rightarrow \boxed{36}.<math></math>
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<cmath>\frac{31}{5} \rightarrow \boxed{36}.</cmath>
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2015|n=II|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:18, 29 March 2015

Problem

Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$. The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

Let us call the quantity $\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \[\frac{ab+1}{a+b} < \frac{3}{2}\] \[\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2\] \[\implies a < \frac{3b - 2}{2b - 3}.\] Now, observe that if $b = 1$ we have that $N = \frac{a + 1}{a + 1} = 1$, regardless of the value of $a$. If $a = 1$, we have the same result: that $N = \frac{b + 1}{b + 1} = 1$, regardless of the value of $b$. Hence, we want to find pairs of positive integers $(a, b)$ existing such that neither $a$ nor $b$ is equal to $1$, and that the conditions given in the problem are satisfied in order to check that the maximum value for $N$ is not $1$.


To avoid the possibility that $a = 1$, we want to find values of $b$ such that $\frac{3b - 2}{2b - 3} > 2$. If we do this, we will have that $a < \frac{3b - 2}{2b - 3} = k$, where $k$ is greater than $2$, and this allows us to choose values of $a$ greater than $1$. Again, since $b$ is a positive integer, and we want $b > 1$, we can legitimately multiply both sides of $\frac{3b - 2}{2b - 3} > 2$ by $2b - 3$ to get $3b - 2 > 4b - 6 \implies b < 4$. For $b = 3$, we have that $a < \frac{7}{3}$, so the only possibility for $a$ greater than $1$ is obviously $2$. Plugging these values into $N$, we have that $N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}$. For $b = 2$, we have that $a < \frac{4}{1} = 4$. Plugging $a = 2$ and $b = 3$ in for $N$ yields the same result of $\frac{31}{5}$, but plugging $a = 2$ and $b = 2$ into $N$ yields that $N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}$. Clearly, $\frac{31}{5}$ is the largest value we can have for $N$, so our answer is $31 + 5 = \boxed{036}$.

Solution 2 (Proof without words)

\[\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b,\] \[\rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.\]

\[2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow\] \[(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\] \[(a, b) = (2, 2), (2, 3), (3, 2).\] \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.\] \[\frac{31}{5} \rightarrow \boxed{36}.\]

See also

2015 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

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