Difference between revisions of "2015 AIME II Problems/Problem 8"
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==Problem== | ==Problem== | ||
− | Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>a</math> and <math>b</math> be positive integers satisfying <math>\frac{ab+1}{a+b} < \frac{3}{2}</math>. The maximum possible value of <math>\frac{a^3b^3+1}{a^3+b^3}</math> is <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
− | ==Solution== | + | ==Solution 1== |
Let us call the quantity <math>\frac{a^3b^3+1}{a^3+b^3}</math> as <math>N</math> for convenience. Knowing that <math>a</math> and <math>b</math> are positive integers, we can legitimately rearrange the given inequality so that <math>a</math> is by itself, which makes it easier to determine the pairs of <math>(a, b)</math> that work. Doing so, we have <cmath>\frac{ab+1}{a+b} < \frac{3}{2}</cmath> <cmath>\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2</cmath> <cmath>\implies a < \frac{3b - 2}{2b - 3}.</cmath> Now, observe that if <math>b = 1</math> we have that <math>N = \frac{a + 1}{a + 1} = 1</math>, regardless of the value of <math>a</math>. If <math>a = 1</math>, we have the same result: that <math>N = \frac{b + 1}{b + 1} = 1</math>, regardless of the value of <math>b</math>. Hence, we want to find pairs of positive integers <math>(a, b)</math> existing such that neither <math>a</math> nor <math>b</math> is equal to <math>1</math>, and that the conditions given in the problem are satisfied in order to check that the maximum value for <math>N</math> is not <math>1</math>. | Let us call the quantity <math>\frac{a^3b^3+1}{a^3+b^3}</math> as <math>N</math> for convenience. Knowing that <math>a</math> and <math>b</math> are positive integers, we can legitimately rearrange the given inequality so that <math>a</math> is by itself, which makes it easier to determine the pairs of <math>(a, b)</math> that work. Doing so, we have <cmath>\frac{ab+1}{a+b} < \frac{3}{2}</cmath> <cmath>\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2</cmath> <cmath>\implies a < \frac{3b - 2}{2b - 3}.</cmath> Now, observe that if <math>b = 1</math> we have that <math>N = \frac{a + 1}{a + 1} = 1</math>, regardless of the value of <math>a</math>. If <math>a = 1</math>, we have the same result: that <math>N = \frac{b + 1}{b + 1} = 1</math>, regardless of the value of <math>b</math>. Hence, we want to find pairs of positive integers <math>(a, b)</math> existing such that neither <math>a</math> nor <math>b</math> is equal to <math>1</math>, and that the conditions given in the problem are satisfied in order to check that the maximum value for <math>N</math> is not <math>1</math>. | ||
− | 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 = | + | 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>. |
− | + | (Technically, we would have to find that b > 1 before dividing both sides of the inequality by 2b - 3, but otherwise this solution is correct) | |
− | |||
− | |||
− | <cmath>2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \} | + | ==Solution 2 (best solution)== |
+ | <cmath>\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.</cmath> | ||
+ | |||
+ | <cmath>2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \} \rightarrow</cmath> <math>(2a-3)</math> and <math>(2b-3)</math> each cannot be even or else <math>a</math> and <math>b</math> will not be integers | ||
<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> | <cmath>(a, b) = (2, 2), (2, 3), (3, 2).</cmath> | ||
<cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.</cmath> | <cmath>\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.</cmath> | ||
− | <cmath>\frac{31}{5} \rightarrow \boxed{ | + | <cmath>\frac{31}{5} \rightarrow \boxed{036}.</cmath> |
+ | |||
+ | (<math>a=1 \rightarrow b=1, 2, 3... \rightarrow \frac{b^3+1}{b^3+1}=1</math>; <math>1<31/5</math>). | ||
+ | |||
+ | ==Solution 3== | ||
+ | Notice that for <math>\frac{a^3b^3+1}{a^3+b^3}</math> to be maximized, <math>\frac{ab+1}{a+b}</math> has to be maximized. We simplify as above to <math>2ab + 2 < 3a + 3b</math>, which is <math>(a-\frac{3}{2})(b-\frac{3}{2}) < \frac{5}{4}</math>. To maximize, <math>a</math> has to be as close to <math>b</math> as possible, making <math>a</math> close to <math>\frac{3+\sqrt{5}}{2}</math>. Because <math>a</math> and <math>b</math> are positive integers, <math>a = 3</math>, and checking back gives <math>b = 2</math> as the maximum or the other way around, which the answer is thus <math>\frac{216+1}{27+8} = \frac{217}{35} = \frac{31}{5} \rightarrow \boxed{036}</math>. | ||
+ | ==Solution 4== | ||
+ | Guess and check a few values of <math>a</math> and <math>b</math> you will find two things. One, that the highest values of <math>a</math> and <math>b</math> are the closest to <math>\frac{3}{2}</math>. Two, that the pair <math>(2,3)</math> is the highest possible value of <math>a, b</math>. So plugging in <math>a=2</math> and <math>b=3</math> we get <math>((2*3)^3)+1/(8+27)</math> = <math>\frac{217}{35}</math> = <math>\frac{31}{5}</math>, and <math>31+5 = 36</math>. | ||
+ | Solution by Helloitsaaryan | ||
==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}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:16, 5 February 2022
Problem
Let and be positive integers satisfying . The maximum possible value of is , where and are relatively prime positive integers. Find .
Solution 1
Let us call the quantity as for convenience. Knowing that and are positive integers, we can legitimately rearrange the given inequality so that is by itself, which makes it easier to determine the pairs of that work. Doing so, we have Now, observe that if we have that , regardless of the value of . If , we have the same result: that , regardless of the value of . Hence, we want to find pairs of positive integers existing such that neither nor is equal to , and that the conditions given in the problem are satisfied in order to check that the maximum value for is not .
To avoid the possibility that , we want to find values of such that . If we do this, we will have that , where is greater than , and this allows us to choose values of greater than . Again, since is a positive integer, and we want , we can legitimately multiply both sides of by to get . For , we have that , so the only possibility for greater than is obviously . Plugging these values into , we have that . For , we have that . Plugging and in for yields the same result of , but plugging and into yields that . Clearly, is the largest value we can have for , so our answer is .
(Technically, we would have to find that b > 1 before dividing both sides of the inequality by 2b - 3, but otherwise this solution is correct)
Solution 2 (best solution)
and each cannot be even or else and will not be integers
(; ).
Solution 3
Notice that for to be maximized, has to be maximized. We simplify as above to , which is . To maximize, has to be as close to as possible, making close to . Because and are positive integers, , and checking back gives as the maximum or the other way around, which the answer is thus .
Solution 4
Guess and check a few values of and you will find two things. One, that the highest values of and are the closest to . Two, that the pair is the highest possible value of . So plugging in and we get = = , and . 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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