I'd like to know answer too. I took bruteforce approach and wrote down all possible values of m:
3 5 7 9 11 13 15 17 19 21 23 25 27 29
then I wrote down some prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29
for (1) we exclude 3, 9, 15, 21, 27. I can see that at least 5 suffices the statement.
for (2) we exclude 5, 15, 25. Divisor 3 suffices the statement.
Any ideas, guys ?
IMO answer D is correct.
GMATPrep Divisibilty
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Feep
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Well, the most common answer to "how many prime divisors" is going be 1, since a lot of those odd numbers up there are prime numbers themselves. So let's look at the ones that aren't: 9, 15, 21, 25, 27. All of these are multiples of three, save for 25, but surprise: 25 only has ONE distinct prime divisor: 5. It just happens to be squared. (1) is sufficient; the answer is always 1.
(2) eliminates 15 and 25, and we can eliminate 9 and 27 since they also only have one distinct prime divisor (3^2, 3^3), but alas! 21 breaks down into 3 and 7. Insufficient.
The answer is A.
(2) eliminates 15 and 25, and we can eliminate 9 and 27 since they also only have one distinct prime divisor (3^2, 3^3), but alas! 21 breaks down into 3 and 7. Insufficient.
The answer is A.
Last edited by Feep on Fri Apr 10, 2009 1:58 am, edited 2 times in total.
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anshulseth
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It is indeed A as told by Feep.
Initially I got C, but as pointed out by Feep, 5 & 25 have only one distinct prime.
Nice question, equally nice explanation.
Initially I got C, but as pointed out by Feep, 5 & 25 have only one distinct prime.
Nice question, equally nice explanation.
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vittalgmat
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It is A.
From stmt 1, we eliminate 15, 21. These are the only two number which have 2 different prime factors. The rest have only 1 prime factor. So sufficient.
Stmt 2: After removing mulitples of 5, we are still left with prime numbers like 3, 7 etc which have only 1 prime factor and 21 which has 2 prime factors.
Hence insufficient.
rgds
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From stmt 1, we eliminate 15, 21. These are the only two number which have 2 different prime factors. The rest have only 1 prime factor. So sufficient.
Stmt 2: After removing mulitples of 5, we are still left with prime numbers like 3, 7 etc which have only 1 prime factor and 21 which has 2 prime factors.
Hence insufficient.
rgds
-V
Honestly I'm not convinced with this statement. Based on what I've read in Princeton Review book any answer counts as sufficient one no matter what the answer is (yes, no, 0, 1 and etc).Feep wrote:Well, the most common answer to "how many prime divisors" is going be 1, since a lot of those odd numbers up there are prime numbers themselves.
So just to rephrase this can I say that you consider (2) insufficient because series [3 7 9 11 13 17 19 21 23 27 29] has a number which is divisible by more that one prime number ? But I don't see how this condition comes from question "m is divisible by how many different positive prime numbers?". It doesn't say "m is divisible by single prime number".Feep wrote:(2) eliminates 15 and 25, and we can eliminate 9 and 27 since they also only have one distinct prime divisor (3^2, 3^3), but alas! 21 breaks down into 3 and 7. Insufficient.
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Feep
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You're missing the point of that statement. To data to be sufficient in this question, we aren't going to be able to find a specific value of m, but instead be able to find that every possible of value of m has the same number of prime divisors, in this case, one. Since most of the possible m's are prime numbers, and therefore have only one divisor, our goal is to try and find other m's that have a DIFFERENT number of prime divisors. If we can, then we do not have sufficient data.expert wrote:Honestly I'm not convinced with this statement. Based on what I've read in Princeton Review book any answer counts as sufficient one no matter what the answer is (yes, no, 0, 1 and etc).Feep wrote:Well, the most common answer to "how many prime divisors" is going be 1, since a lot of those odd numbers up there are prime numbers themselves.
So just to rephrase this can I say that you consider (2) insufficient because series [3 7 9 11 13 17 19 21 23 27 29] has a number which is divisible by more that one prime number ? But I don't see how this condition comes from question "m is divisible by how many different positive prime numbers?". It doesn't say "m is divisible by single prime number".Feep wrote:(2) eliminates 15 and 25, and we can eliminate 9 and 27 since they also only have one distinct prime divisor (3^2, 3^3), but alas! 21 breaks down into 3 and 7. Insufficient.
(2) is insufficient because 21 is included as a possibility, which has two prime divisors, whereas every other possible value of m has one.
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