We want to find the number of positive integers not exceeding 1000 that are either the square or the cube of an integer.
We will get that there are 38 of these.
First, the squares:
To get the number of square numbers not exceeding 1000, we need to find the largest integer that when squared is equal or smaller than 1000.
This is given by:
√1000 = 31.6
Then the largest square number smaller than 1000 is given by:
31^2 = 961
From this, we can conclude that there are 31 integers not exceeding 100 that are the square of an integer.
For the cubes we do the same thing:
∛1000 = 10
The largest cube number that meets the condition is:
10^3 = 1000
Then there are 10 cube numbers that meet the condition.
Now we need to find the common numbers in bot groups (this is, numbers that are a cube and a square of two integers and that are smaller than 1000).
The numbers that meet this condition are:
1)
1^2 = 1
1^3 = 1
64)
4^3 = 64
8^2 = 64
81)
9^3 = 279
27^2 = 729
These 3 numbers are in both groups, so we are counting them twice, thus we need to subtract 3 to the final result, in order to avoid counting the same number two times.
Then the total number of positive integers not exceeding 1000 that are either the square or the cube of an integer is:
31 + 10 - 3 = 38
If you want to learn more, you can read:
https://brainly.com/question/1571795
