Hi All,

These types of questions don't require a complex math approach - they can be solved with a bit of 'brute force' and logic.

In real basic terms, we're asked to find all of the '3s' in 200!

We can figure out that 200/3 = 66, so we know that there are at least 66 '3s' in 200! While that answer is among the 5 choices, it seems a bit too 'easy', so let's do a bit more work and list out the first few numbers that we know have a '3' in them:

3 = 3x1

6 = 3x2

9 = 3x3

Notice how both 3 and 6 have just one 3 in them, but 9 has TWO 3s (there's an 'extra' 3 that we have to account for). This implies that there are probably other numbers that include 'extra 3s' that we have to figure out:

To find those extra 3s, we have to look at numbers that contain 'powers of 3'...

3^2 = 9

3^3 = 27

3^4 = 81

3^5 = 243, but that's too big (we're only going up to 200). Keep in mind that a multiple of 81 is also a multiple of 9 and 27, so we don't want to count any of those values more than once.

200/9 = 22, so we know that there are at least 22 extra 3s (and certainly more because of the 27 and 81). With the 66 3s that we already have, those 22 extra 3s increase the total to 88. With the other extra 3s, we'll end up with MORE than 88 3s. There's only one answer that fits that logic...

Final Answer:

GMAT assassins aren't born, they're made,

Rich

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# Rich Cohen

Co-Founder & GMAT Assassin

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