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Difference between revisions of "2015 AIME II Problems/Problem 8"
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Revision as of 21:04, 9 August 2020

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 (best solution)

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

\[2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \} \rightarrow\] $(2a-3)$ and $(2b-3)$ each cannot be even or else $a$ and $b$ will not be integers \[(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\] \[(a, b) = (2, 2), (2, 3), (3, 2).\] \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.\] \[\frac{31}{5} \rightarrow \boxed{036}.\]

($a=1 \rightarrow b=1, 2, 3... \rightarrow \frac{b^3+1}{b^3+1}=1$; $1<31/5$).

Solution 3

Notice that for $\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\frac{ab+1}{a+b}$ 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 or the other way around, which the answer is thus $\frac{216+1}{27+8} = \frac{217}{35} = \frac{31}{5} \rightarrow \boxed{036}$.

Solution 4

Guess and check a few values of $a$ and $b$ you will find two things. One, that the highest values of $a$ and $b$ are the closest to $\frac{3}{2}$. Two, that the pair $(2,3)$ is the highest possible value of $a, b$. So plugging in $a=2$ and $b=3$ we get $((2*3)^3)+1/(8+27)$ = $\frac{217}{35}$ = $\frac{31}{5}$, and $31+5 = 36$. Solution by Helloitsaaryan

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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