Difference between revisions of "1990 AIME Problems/Problem 1"
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The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neither the [[perfect square | square]] nor the [[perfect cube | cube]] of a positive integer. Find the 500th term of this sequence. | The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neither the [[perfect square | square]] nor the [[perfect cube | cube]] of a positive integer. Find the 500th term of this sequence. | ||
− | == Solution == | + | == Solution 1== |
− | Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than <math>500</math>. This happens to be <math>23^2=529</math>. Notice that there are <math>23</math> squares and <math>8</math> cubes less than or equal to <math>529</math>, but <math>1</math> and <math>2^6</math> are both squares and cubes. Thus, there are <math>529-23-8+2=500</math> numbers in our sequence less than <math>529</math>. Magically, we want the <math>500th</math> term, so our answer is the | + | Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than <math>500</math>. This happens to be <math>23^2=529</math>. Notice that there are <math>23</math> squares and <math>8</math> cubes less than or equal to <math>529</math>, but <math>1</math> and <math>2^6</math> are both squares and cubes. Thus, there are <math>529-23-8+2=500</math> numbers in our sequence less than <math>529</math>. Magically, we want the <math>500th</math> term, so our answer is the biggest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>. |
− | <math></math> | + | |
+ | == Solution 2== | ||
+ | This solution is similar as Solution 1, but to get the intuition why we chose to consider <math>23^2 = 529</math>, consider this: | ||
+ | |||
+ | We need <math>n - T = 500</math>, where <math>n</math> is an integer greater than 500 and <math>T</math> is the set of numbers which contains all <math>k^2,k^3\le 500</math>. | ||
+ | |||
+ | Firstly, we clearly need <math>n > 500</math>, so we substitute n for the smallest square or cube greater than <math>500</math>. However, if we use <math>n=8^3=512</math>, the number of terms in <math>T</math> will exceed <math>n-500</math>. Therefore, <math>n=23^2=529</math>, and the number of terms in <math>T</math> is <math>23+8-2=29</math> by the [[Principle of Inclusion-Exclusion]], fulfilling our original requirement of <math>n-T=500</math>. | ||
+ | As a result, our answer is <math>529-1 = \boxed{528}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1990|before=First Question|num-a=2}} | {{AIME box|year=1990|before=First Question|num-a=2}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:11, 25 June 2023
Contents
Problem
The increasing sequence consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence.
Solution 1
Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than . This happens to be
. Notice that there are
squares and
cubes less than or equal to
, but
and
are both squares and cubes. Thus, there are
numbers in our sequence less than
. Magically, we want the
term, so our answer is the biggest non-square and non-cube less than
, which is
.
Solution 2
This solution is similar as Solution 1, but to get the intuition why we chose to consider , consider this:
We need , where
is an integer greater than 500 and
is the set of numbers which contains all
.
Firstly, we clearly need , so we substitute n for the smallest square or cube greater than
. However, if we use
, the number of terms in
will exceed
. Therefore,
, and the number of terms in
is
by the Principle of Inclusion-Exclusion, fulfilling our original requirement of
.
As a result, our answer is
.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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.