IanStewart wrote:
Splitopia wrote:
Out of curiosity why wouldn't you need to know the number of primes less than 122?
What Bunuel means is: the GMAT will never test if you know how many primes there are less than 122. That's definitely true. So you'd never need to know that if you're taking the GMAT. If instead we're discussing how to answer this particular question, then you certainly need to know, or figure out, how many primes there are less than 122, but the question in this thread is not a realistic GMAT question.
A couple of posters above suggest it's worth knowing there are 25 primes less than 100. I can't imagine the GMAT asking a question where that knowledge would be helpful, and I'd be curious if anyone can point to a single official GMAT question where you'd benefit from knowing how many primes there are less than 100.
Thanks for clarifying. The GMAT seems to have a lot of practice questions out there that don't conform to the way the test challenges test takers. From most official answers I've read, the difficulty is in determining the approach, the underlying relations, what they're really asking, or how we could go about testing something. A lot of practice questions miss the mark by making the GMAT more about algebra or brute force than finesse and reasoning. Still helpful to know I won't see something like this come test day.
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On topic I also considered another approach for this question before I knew I wasn't likely to come across it again. This is without the 25 primes in 1-100 knowledge.
-> 3 factors therefore only counting squares of primes.
-> sqrt(15000) = sqrt(100 150) = sqrt(100 25 6) = 50sqrt(6). To test for sqrt 6 I considered sqrt(600). 24 shoots under (144*4 = 576) and 25 shoots over (5^4 = 625). It's almost exactly half so 24.5 should be a good approximation. Which makes 2.45 close to sqrt(6) value.
50*2.45 = 122.5 So we're looking for all the primes up to 122.
I first take out 1 (-1) as we won't catch it later. Then I take out non prime numbers with factors of 2 or 3.
122/2 v = 61. But the original 2 is a prime so it's (-60)
122/3 v = 40. But the original 3 is a prime so it's (-39)
Add back sixes as we removed these twice. 122/6 = (+20)
We could do the same for say 5, but it would get complicated. Ex: We'd end up removing 90 again by removing multiples of 15. It gets too messy to add back sub groups. Instead we can note any multiple of a larger prime has not been removed if that multiple is also a prime higher than 2 or 3.
That is to say 5* 2/3/4/6/8/9/10 etc is accounted for. but 5 * prime that isn't 2/3 has not been accounted for.
So let's remove what we haven't handled. 122/5 v = 24 so we are looking for primes up to 24.
5*5, 5*7, 5*11, 5*13, 5*17, 5*19, 5*23. Don't need to solve, we know they aren't double counted.
Same logic for larger primes. 122/7 v = 17. 122/11 v = 11. 122/13 = 9 (9 is the highest factor that fits, everything has been accounted for)
7*7, 7*11, 7*13, 7*17
11*11
(-12)
122 - 1 - 60 - 39 + 20 - 12 = 30
Probably not worth brute forcing, but anyways it's a different way to approach I guess.
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