If , and where denotes the greatest integer less than or equal to x, then the number of possible values of x will be
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This one took me a while. The key is realizing the fractional part terms have to vanish, right? That only happens when x is divisible by all of 2, 3, and 5. Am I on the right track?
Yes, that's the trick. Once you see that, it's just a counting problem up to 1000.
Wait, why can't x be like 7? The equation would still hold approximately... Oh I see, the floors make it exact, so any remainder breaks it. Clever problem.
Yeah, the equality is exact so any leftover fraction messes it up. Took me a minute too.
Counted multiples of 30 below 1000 and got 33, but the answer choices include both 33 and 44/45. Did anyone else second-guess whether 0 counts? x>0 so no.
Same trap here! I almost counted 0 because 0 is divisible by everything.
Good 655-705 level question. Not conceptually hard but easy to misread the inequality. Respect.
What it tests
Your ability to evaluate and compose functions, including unusual defined operations.
Common trap
Applying function operations in the wrong order or ignoring domain restrictions.