1990 AIME Problems/Problem 1
Contents
[hide]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) | ||
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