The list A is the list of all positive integers from 1 to 1000. We create a list B by changing all the multiples of 3 from the list A by its successors. Then we create the list C by changing all the multiples of 7 from the list B by its predecessors.

How many numbers in list A still appears at least one time in the list C?

Guest Jul 2, 2020

#1**+1 **

Yay I'm back after months of video games and coding.

What DOESN'T appear one time? We can call that being "deleted".

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Let's call a term in list A "\(n\)"

- \(n\) is only deleted if \(n\text{ }mod\text{ }4=3 \) and not \(n + 1\text{ }mod\text{ }8=7\)

so in other words, if \(n\) is a multiple of 3 and \(n+1\) is NOT a multiple of 7, then it shall be deleted.

Example {n = 3}

- n is multiple of 3

- n + 1 is NOT multiple of 7

therefore it should be deleted.

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Now let us calculate how much will be deleted.

There are floor(1000/3) = 333 multiples of 3 from 1 - 1000.

There are floor(333/7) = 47 multiples of 7 that are also a multiple of 3.

That means there are 333 - 47 = 286 multiples of 3 that are NOT multiples of 7. That means 286 should be deleted.

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Since the question asks how many numbers are not deleted in list C, we can easily find that the answer is 1000 - 286 is \(\boxed{714}\).

Please correct me if I am wrong.

AnExtremelyLongName Jul 2, 2020