Difference between revisions of "1990 AIME Problems/Problem 1"
Danielguo94 (talk | contribs) (→Solution) |
Danielguo94 (talk | contribs) (→Solution) |
||
| Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
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 smallest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>. | 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 smallest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>. | ||
| − | <math></math> | + | <math></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]] | ||
Revision as of 23:57, 7 September 2011
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
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 smallest non-square and non-cube less than , which is 
.
$$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
