Difference between revisions of "1987 AIME Problems/Problem 12"
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== Problem == | == Problem == | ||
Let <math>m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>n+r</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. | Let <math>m</math> be the smallest [[integer]] whose [[cube root]] is of the form <math>n+r</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. | ||
− | == Solution == | + | == Solution 1 == |
In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible. | In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible. | ||
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In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>. The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>. Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set. The answer is <math>019</math>. | In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>. The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>. Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set. The answer is <math>019</math>. | ||
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+ | == Solution 2 == | ||
+ | We have that <math>(n + 1/1000)^3 = n^3 + 3/10^3 n^2 + 3/10^6 n + 1/10^9</math>. If <math>(n + 1/1000)^3 - n^3 > 1</math>, there exists an integer between <math>(n + 1/1000)^3</math> and <math>n^3</math>, and the cube root of this integer would be between <math>n</math> and <math>n + 1/1000</math>. We seek the smallest <math>n</math> such that <math>(n + 1/1000)^3 - n^3 > 1</math>. | ||
+ | |||
+ | <cmath>(n + 1/1000)^3 - n^3 > 1</cmath> | ||
+ | <cmath>3/10^3 n^2 + 3/10^6 n + 1/10^9 > 1</cmath> | ||
+ | <cmath>3n^2 + 3/10^3 n + 1/10^6 > 10^3</cmath>. | ||
+ | |||
+ | Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</math>. | ||
== See also == | == See also == |
Revision as of 13:52, 23 March 2018
Contents
[hide]Problem
Let be the smallest integer whose cube root is of the form
, where
is a positive integer and
is a positive real number less than
. Find
.
Solution 1
In order to keep as small as possible, we need to make
as small as possible.
. Since
and
is an integer, we must have that
. This means that the smallest possible
should be quite a bit smaller than 1000. In particular,
should be less than 1, so
and
.
, so we must have
. Since we want to minimize
, we take
. Then for any positive value of
,
, so it is possible for
to be less than
. However, we still have to make sure a sufficiently small
exists.
In light of the equation , we need to choose
as small as possible to ensure a small enough
. The smallest possible value for
is 1, when
. Then for this value of
,
, and we're set. The answer is
.
Solution 2
We have that . If
, there exists an integer between
and
, and the cube root of this integer would be between
and
. We seek the smallest
such that
.
.
Trying values of , we see that the smallest value of
that works is
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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