the positive integer k has exactly two positive prime factors, 3 and 7. If K has a total of 6 positive factors, including 1 and k, what is the value of K?
1) 3^2 is a factor of k
2) 7^2 is NOT a factor of k
OA: D
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- shobuj40
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- JonathanSchneider
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Re: the positive integer k has
Where did you get this question from? I don't recognize it, and the wording seems a bit strange. Did you make it up? Also, do you have any specific questions about it?
- gauravbawa+gm
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Re: the positive integer k has
Since it has only 2 prime factors but 6 factors (4 of which are 1, 3, 7, k) this means that the prime factors must be combined to generate the other 2 factors - the other 2 can only be either 3 which means 3x3=9 and 3.7=21 is a factor OR 7 which means the other 2 factors are 21 and 49.
- RonPurewal
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Re: the positive integer k has
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).
--
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.
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).
--
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.
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