by RonPurewal Fri Feb 27, 2009 7:52 am
SHORTCUT METHOD:
if you know the following useful fact, then you can solve this problem much more quickly.
USEFUL FACT: if a, b, ... are the EXPONENTS in the prime factorization of a number, then the total number of factors of that number is the product of (a + 1), (b + 1), ...
example:
540 = (2^2)(3^3)(5^1), in which the exponents are 2, 3, and 1. therefore, 540 has (2 + 1)(3 + 1)(1 + 1) = 3 x 4 x 2 = 24 different factors.
with this shortcut method, realize that 6 (the total number of factors) is 3 x 2. therefore, the exponents in the prime factorization must be 2 and 1, in some order.
therefore, there are only two possibilities: k = (3^2)(7^1) = 63, or k = (3^1)(7^2) = 147.
statement (1) includes 63 but rules out 149, so, sufficient.
statement (2) includes 63 but rules out 149, so, sufficient.
answer = (d).
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IF YOU DON'T KNOW THE SHORTCUT:
statement (1)
if 3^2 is a factor of k, then so is 3^1.
therefore, we already have four factors: 1, 3^1, 3^2, and 7.
but we also know that (3^1)(7) and (3^2)(7) must be factors, since 3^2 and 7 are both part of the prime factorization of k.
that's already six factors, so we're done: k must be (3^2)(7). if it were any bigger, then there would be more than these six factors.
sufficient.
statement (2)
if 7 is a factor of k, but 7^2 isn't, then the prime factorization of k contains EXACTLY one 7.
therefore, we need to find out how many 3's will produce six factors when paired with exactly one 7.
in fact, it's data sufficiency, so we don't even have to find this number; all we have to do is realize that adding more 3's will always increase the number of factors, so, there must be exactly one number of 3's that will produce the correct number of factors. (as already noted above, that's two 3's, or 3^2.)
sufficient.