Hey, nice explanation on this one. Just to expand that a little bit more with some strategy, because this question asks for a product you know that the the numbers 2, 3, 5, and 7 are being multiplied together in different quantities of each. The easiest number to factor out is 10, which is 2*5 - the number of zeroes in this problem will tell us exactly how many tens it is divisible by. There are 3 zeroes, so we know that this number is:
10^3 * 756
From there, 756 isn't as bad to factor out. because it's even you know that you can divide by 2:
10^3 * 2 * 378
And it's still even so you can do it again:
10^3 * 2 * 2 * 189
And now it's no longer even, but you can tell that it's divisible by 9 (the sum of the digits is 18), so you can call it:
10^3 * 2 * 2 * 21 * 9
And then factor out the 21 and the 9:
10^3 * 2 * 2 * 3 * 7 * 3 * 3
The question asks for the number of 3s, so count those up and there are 3 of them, so the answer is C.
Strategically here, just start with the easiest numbers to factor out and work from there. 10s and 2s are usually the easiest - 10s just become zeroes at the end, and you can usually divide by 2 in your head, so you can make fairly easy work of these by starting with easy factors and working from there.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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