2015 AIME II Problems/Problem 8

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 1

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 $a = 3$ and $b = 2$ 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)

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

$2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \}; \rightarrow$ $(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).$ $(a, b) = (2, 2), (2, 3), (3, 2).$ $\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.$ $\frac{31}{5} \rightarrow \boxed{036}.$

Solution 2

Notice that for $\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\frac{ab+1}{ab}$ has to be maximized. We simplify as above to $2ab + 2 < 3a + 3b$ , which is $(a-\frac{3}{2})(b-\frac{3}{2}) < \frac{5}{4}$ . To maximize, $a$ has to be as close to $b$ as possible, making $a$ close to $\frac{3+\sqrt{5}}{2}$ . Because $a$ and $b$ are positive integers, $a = 3$ , and checking back gives $b = 2$ as the maximum, which the answer is thus $\frac{216+1}{27+8} = \frac{217}{35} = \frac{31}{5} \rightarrow \boxed{036}$ .

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2015 AIME II Problems/Problem 8

Contents

Problem

Solution 1

Solution 2 (Proof without words)

Solution 2

See also