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heheh... really dont know wether my thinking is correct..
I got to 66 as well using two strategies... the first one is the same as GSR. The second is:

prime factorization of all 10 is 2*5. The next integer divisable by 5 would be 15... thus 3*5... next one 4*5... etc till 30*5 = 150.
The number of integers divisble by 5 from 10 to 100 inclusive then is:
30 - 2 + 1 = 29

Same for 3.. so the first number after 11 to be divisable by 3 is 12. that is 2*2*3 or 4*3... next one is 5*3 ... etc till 50*3 = 150.
The number of integers divisible by 3 from 10 to 100 inclusive then is:
50 - 4 + 1 = 47

Now we add these two numbers 29 + 47 = 76. BUT, we have to substract 1 because 3*5 and 5*3 has been counted in both. so 75.

Total integers between 10 and 150 inclusive is 150 - 10 + 1 = 141
So total integers NOT divisible by 5 nor 3 is 141 - 75 = 66

NOW , since the question is asking for evenly divisable integers, and the prime factorization demonstrated that the integers are "consecutive", we can assume that every second number will result an even integer. Thus 66/2 = 33 -> A

I am extremely sorry guys. I didn't realize that I have typed 10 instead of 100, even during my first edit

I really apologize for straining your brains unnecessarily. I have edited the question and it should be correct now. I would post the OA by end of today.

Numbers between 100 and 150 inclusive = 51 (includes 100 AND 150, or does the inclusive only apply to the 150??)
Numbers divisible by 5 = (150 - 100)/5 + 1 = 11
Numbers divisible by 3 = (150-102)/3 + 1 = 17
Numbers divisible by 3&5 = 105, 120, 135, 150 = 4

lol, I saw this on the forum yesterday and posted a note saying that I thought the answer should be C (27). I got home and cracked open my Princeton Review book for some practice and there was the same problem. I worked it out again and got 27. OA was indeed C (27) in the Princeton Review book.

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