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

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To avoid the possibility that <math>a = 1</math>, we want to find values of <math>b</math> such that <math>\frac{3b - 2}{2b - 3} > 2</math>. If we do this, we will have that <math>a < \frac{3b - 2}{2b - 3} = k</math>, where <math>k</math> is greater than <math>2</math>, and this allows us to choose values of <math>a</math> greater than <math>1</math>. Again, since <math>b</math> is a positive integer, and we want <math>b > 1</math>, we can legitimately multiply both sides of <math>\frac{3b - 2}{2b - 3} > 2</math> by <math>2b - 3</math> to get <math>3b - 2 > 4b - 6 \implies b < 4</math>. For <math>b = 3</math>, we have that <math>a < \frac{7}{3}</math>, so the only possibility for <math>a</math> greater than <math>1</math> is obviously <math>2</math>. Plugging these values into <math>N</math>, we have that <math>N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}</math>. For <math>b = 2</math>, we have that <math>a < \frac{4}{1} = 4</math>. Plugging <math>a = 2</math> and <math>b = 3</math> in for <math>N</math> yields the same result of <math>\frac{31}{5}</math>, but plugging <math>a = 2</math> and <math>b = 2</math> into <math>N</math> yields that <math>N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}</math>. Clearly, <math>\frac{31}{5}</math> is the largest value we can have for <math>N</math>, so our answer is <math>31 + 5 = \boxed{036}</math>.
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To avoid the possibility that <math>a = 1</math>, we want to find values of <math>b</math> such that <math>\frac{3b - 2}{2b - 3} > 2</math>. If we do this, we will have that <math>a < \frac{3b - 2}{2b - 3} = k</math>, where <math>k</math> is greater than <math>2</math>, and this allows us to choose values of <math>a</math> greater than <math>1</math>. Again, since <math>b</math> is a positive integer, and we want <math>b > 1</math>, we can legitimately multiply both sides of <math>\frac{3b - 2}{2b - 3} > 2</math> by <math>2b - 3</math> to get <math>3b - 2 > 4b - 6 \implies b < 4</math>. For <math>b = 3</math>, we have that <math>a < \frac{7}{3}</math>, so the only possibility for <math>a</math> greater than <math>1</math> is obviously <math>2</math>. Plugging these values into <math>N</math>, we have that <math>N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}</math>. For <math>b = 2</math>, we have that <math>a < \frac{4}{1} = 4</math>. Plugging <math>a = 3</math> and <math>b = 2</math> in for <math>N</math> yields the same result of <math>\frac{31}{5}</math>, but plugging <math>a = 2</math> and <math>b = 2</math> into <math>N</math> yields that <math>N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}</math>. Clearly, <math>\frac{31}{5}</math> is the largest value we can have for <math>N</math>, so our answer is <math>31 + 5 = \boxed{036}</math>.
  
 
==Solution 2 (Proof without words)==
 
==Solution 2 (Proof without words)==

Revision as of 21:45, 7 July 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 $a = 3$ 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)

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

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

See also

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