Difference between revisions of "2015 AIME II Problems/Problem 8"
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==Solution 1== | ==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^3 + 1}{a^3 + 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^3 + 1}{b^3 + 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 = 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>. | 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 \} \rightarrow</cmath> | + | ==Solution 2 (nifty 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> | ||
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<cmath>\frac{31}{5} \rightarrow \boxed{036}.</cmath> | <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== | ==Solution 3== | ||
− | Notice that for <math>\frac{a^3b^3+1}{a^3+b^3}</math> to be maximized, <math>\frac{ab+1}{ | + | 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 | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=9re2qLzOKWk&t=653s | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==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 22:39, 14 February 2024
Contents
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 (nifty 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
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=653s
~MathProblemSolvingSkills.com
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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