To start \(1000=(5*2)^3\), and 1000 has two factors 5 and 2. So, by excluding the set of all even numbers which are \(1000-500=500\) and the set of all multplies of 5, \(\frac{1000}{5}=200\) and adding back the overlap between two sets of 2 and 5, \(\frac{1000}{5*2}=100\), the answer will be \(1000-500-200+100=400\) and it's C

This one is tricky since the Set N contains all integers \(x\) such that x < 1000, including negative integers and 0.

This might be why the answer is given as D, as this could be interpreted differently. Though if taken literally, I would assume A since we are essentially removing all multiples of 2 or 5 from the set of all integers defined by \(∞ < x < 1000\)

On the test I'd assume they'd put the set of all positive integers, leading to motion's solution.

c

\(1000 = 2^3*5^3\)

So basically all even integers are automatically eliminated

There are 500 odd integers from 0 to 1000

Now we also have 5 as a factor. Multiples of 5 have either 0 or 5 at their one's digit place.
The ones with 0 at one's digit place are already eliminated since they are even.

We have to eliminate all integers that have 5 at their one's digit place. There are 100 such numbers.

Therefore, \(N = 500 - 100 = 400\)

But, the question mentions only integer, there is nothing given about it needing to be positive or negative. Even if you don't consider negative number, 0 satisfies the questions requirements and now \(N\) = 401

Considering negative numbers will practically makes \(N = ∞\) since you can consider integers up to \(-∞\)

Hence, Answer is A

I'm confused. The question doesn't specify prime factors, so wouldn't numbers with factors of 4, 5, 8, and 10 also be included?

The question as pointed out above should be worded better
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To find the integers with non-common factors with 1000, let's first find the integers which do have factors in common with 1000.

To find that, there will be 3 types of integers, single digits, 2 digits and 3 digits.
for single digit numbers, the possible choices are - 0,2,4,6,8,5 hence 6C1.
for 2 digit numbers, 9C1*6C1
for 3 digit numbers, 9C1*10C1*6C1

Adding them all, 6+54+540=600

Hence the numbers which do not have common factors with 1000 = 1000-600 = 400
Hence, the answer should be C.