Question 772189: Professor,
Not sure as to if this is the right section because I am in my 60s and haven't had anyone ask me a question like this:
How many integers from 1 to 1000 (exclusive) are divisible by the cube of an integer larger than or equal to 2
If you are as wise as an owl and can help me solve this for a friend please send me the solution as to how you came up with this answer and answer.
Regards,
snowhawk@mia.net
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Since 10 cubed IS 1000 and 1000 is excluded, we are only concerned with the cubes from 2 through 9, namely 8, 27, 64, 125, 216, 343, 512, and 729.
Of the base numbers, 2 through 9, we only need consider the primes, namely 2, 3, 5, and 7, and therefore the cubes 8, 27, 125, and 343. That is because if you count the cubes of the composites, you will be double or triple counting some of them.
2 cubed is 8, 8 divides 1000 exactly 125 times. Exclude one because 1000 is excluded, hence 124 integers between 1 and 1000 that are divisible by 8.
3 cubed is 27. 27 divides 1000 a fraction more than 37 times. Round down. 37 integers divisible by 27.
5 cubed is 125. 125 divides 1000 exactly 8 times. Exclude one because 1000 is excluded, hence 7 integers between 1 and 1000 that are divisible by 125.
7 cubed is 343. 343 divides 1000 a fraction less than 3. Round down. 2 integers between 1 and 1000 divisible by 343.
All together: 124 plus 37 plus 7 plus 2 is 170 integers between 1 and 1000 exclusive that are divisible by cubes of integers greater than or equal to 2.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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