Difference between revisions of "2015 AIME II Problems/Problem 8"
Expilncalc (talk | contribs) (→Solution 2 (Proof without words): Case where a, b = 1) |
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<cmath>\frac{31}{5} \rightarrow \boxed{036}.</cmath> | <cmath>\frac{31}{5} \rightarrow \boxed{036}.</cmath> | ||
− | Notice that if <math>a</math> were 1, giving <math>-1</math> | + | Notice that if <math>a</math> were 1, giving <math>-1</math> for <math>2a-3</math>, the fraction would just be <math>\frac{b^3+1}{b^3+1}=1</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 09:31, 29 August 2017
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
.
Solution 2 (Proof without words)
Notice that if were 1, giving
for
, the fraction would just be
.
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, which the answer is thus
.
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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