Difference between revisions of "2013 AIME I Problems/Problem 8"
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==Operation Quadratics (Solution 3)== | ==Operation Quadratics (Solution 3)== | ||
− | Note that we need <math>-1 | + | Note that we need <math>-1\le f(x)\le 1</math>, and this eventually gets to <math>\frac{m^2-1}{mn}=\frac{1}{2013}</math>. From there, break out the quadratic formula and note that <math>m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}</math>. Then we realize that the square root, call it <math>a</math>, must be an integer. Then <math>(a-n)(a+n)=4026^2.</math> |
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+ | Observe carefully that <math>4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61</math>! It is not difficult to see that to minimize the sum, we want to minimize <math>n</math> as much as possible. Seeing that <math>2a</math> is even, we note that a <math>2</math> belongs in each factor. Now, since we want to minimize <math>a</math> to minimize <math>n</math>, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of <math>2, 61, 61</math> and <math>2, 11, 3, 3, 11</math> fails; the next best is <math>2, 61, 11, 3, 3</math> and <math>2, 61, 11</math>, in which <math>a=6710</math> and <math>n=5368</math>. That is our best solution, upon which we see that <math>m=3</math>, thus <math>\boxed{371}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=7|num-a=9}} | {{AIME box|year=2013|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:57, 15 October 2020
Problem 8
The domain of the function is a closed interval of length
, where
and
are positive integers and
. Find the remainder when the smallest possible sum
is divided by 1000.
Solution
We know that the domain of is
, so
. Now we can apply the definition of logarithms:
Since the domain of
has length
, we have that
A larger value of will also result in a larger value of
since
meaning
and
increase about linearly for large
and
. So we want to find the smallest value of
that also results in an integer value of
. The problem states that
. Thus, first we try
:
Now, we try
:
Since
is the smallest value of
that results in an integral
value, we have minimized
, which is
.
Solution 2
We start with the same method as above. The domain of the arcsin function is , so
.
For to be an integer,
must divide
, and
. To minimize
,
should be as small as possible because increasing
will decrease
, the amount you are subtracting, and increase
, the amount you are adding; this also leads to a small
which clearly minimizes
.
We let equal
, the smallest factor of
that isn't
. Then we have
, so the answer is
.
Operation Quadratics (Solution 3)
Note that we need , and this eventually gets to
. From there, break out the quadratic formula and note that
. Then we realize that the square root, call it
, must be an integer. Then
Observe carefully that ! It is not difficult to see that to minimize the sum, we want to minimize
as much as possible. Seeing that
is even, we note that a
belongs in each factor. Now, since we want to minimize
to minimize
, we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of
and
fails; the next best is
and
, in which
and
. That is our best solution, upon which we see that
, thus
.
See also
2013 AIME I (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.