Difference between revisions of "1990 AIME Problems/Problem 1"
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== 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>. | ||
| + | |||
| + | == Solution 2== | ||
| + | similar as above, but to get the intuition why we chose to consider 23^2 = 529 , consider this | ||
| + | |||
| + | we need n - T = 500, where n = #integers in the list 1,2,..,n and T is the set of numbers which are either k^2 or k^3 and <=n | ||
| + | firstly, we clearly need n > 500 | ||
| + | so we think of taking the smallest square greater than 500 and let that be equal to n( u could try letting n = 512 = 8^3 to with similiar logic, but quickly realise that it fails). This is done so that set T is easy to calculate | ||
| + | so n = 529, set T = 23+8-2 by PIE hence n-T = 500 | ||
| + | so our answer is 529-1 = 528 | ||
== See also == | == See also == | ||
Revision as of 21:29, 27 June 2022
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
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 
.
Solution 2
similar as above, but to get the intuition why we chose to consider 23^2 = 529 , consider this
we need n - T = 500, where n = #integers in the list 1,2,..,n and T is the set of numbers which are either k^2 or k^3 and <=n firstly, we clearly need n > 500 so we think of taking the smallest square greater than 500 and let that be equal to n( u could try letting n = 512 = 8^3 to with similiar logic, but quickly realise that it fails). This is done so that set T is easy to calculate so n = 529, set T = 23+8-2 by PIE hence n-T = 500 so our answer is 529-1 = 528
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. 
