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A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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25 Jan 2014, 07:36
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A Pierpont prime is any prime number p such that \(p =(2^k)(3^l)+1\), where k and l are nonnegative integers. If r is an integer, is r a Pierpont prime? (1) 1 < r < 5 (2) 0 < r < 4
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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25 Jan 2014, 07:55
guerrero25 wrote: A Pierpont prime is any prime number p such that p =(2^k)(3^l)+1 , where k and l are nonnegative integers. If r is an integer, is r a Pierpont prime?
A) 1 < r < 5 B)0 < r < 4
I do not have OA with me right now . Hi, A) 1<r<5 Since r is an integer , then the possible values that can take r are : 2,3 and 4. r is a Pierpont prime if r = (2^k)(3^l)+1 r=2 > r is a Pierpont prime since 2 is prime and 2 can be written as : 2 = (2^0)(3^0) + 1 r=3 > r is a Pierpont prime since 3 is prime and 3 can be written as : 3 = (2^1)(3^0) + 1 r=4 > r is NOT a Pierpont prime since 4 is not prime. Hence, This statement alone is Insufficient B) 0<r<4 Since r is an integer, then the possible values that can take r are : 1,2 and 3 AS seen in Statement 1 , 2 and 3 are Pierpont prime but 1 is not prime , hence this statement is insufficent itself A+B) Now, Statements combined, we should have : 1<r<4 that give r the only two possible values : 2 and 3 and both of them are Pierpont prime as seen before Hence , the answer is Yes . Answer : C
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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10 Aug 2014, 00:53
Rock750 wrote: guerrero25 wrote: A Pierpont prime is any prime number p such that p =(2^k)(3^l)+1 , where k and l are nonnegative integers. If r is an integer, is r a Pierpont prime?
A) 1 < r < 5 B)0 < r < 4
I do not have OA with me right now . Hi, A) 1<r<5 Since r is an integer , then the possible values that can take r are : 2,3 and 4. r is a Pierpont prime if r = (2^k)(3^l)+1 r=2 > r is a Pierpont prime since 2 is prime and 2 can be written as : 2 = (2^0)(3^0) + 1 r=3 > r is a Pierpont prime since 3 is prime and 3 can be written as : 3 = (2^1)(3^0) + 1 r=4 > r is NOT a Pierpont prime since 4 is not prime. Hence, This statement alone is Insufficient B) 0<r<4 Since r is an integer, then the possible values that can take r are : 1,2 and 3 AS seen in Statement 1 , 2 and 3 are Pierpont prime but 1 is not prime , hence this statement is insufficent itself A+B) Now, Statements combined, we should have : 1<r<4 that give r the only two possible values : 2 and 3 and both of them are Pierpont prime as seen before Hence , the answer is Yes . Answer : C Hi, Can anyone explain, how can we have 1 as a solution using statement 2, since 2^0 X 3^0 + 1 = 2, so we can never reach 1. So only options left are 2 & 3 so ans should be statement B.



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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14 Aug 2014, 12:32
PuneetSood wrote: Rock750 wrote: guerrero25 wrote: A Pierpont prime is any prime number p such that p =(2^k)(3^l)+1 , where k and l are nonnegative integers. If r is an integer, is r a Pierpont prime?
A) 1 < r < 5 B)0 < r < 4
I do not have OA with me right now . Hi, A) 1<r<5 Since r is an integer , then the possible values that can take r are : 2,3 and 4. r is a Pierpont prime if r = (2^k)(3^l)+1 r=2 > r is a Pierpont prime since 2 is prime and 2 can be written as : 2 = (2^0)(3^0) + 1 r=3 > r is a Pierpont prime since 3 is prime and 3 can be written as : 3 = (2^1)(3^0) + 1 r=4 > r is NOT a Pierpont prime since 4 is not prime. Hence, This statement alone is Insufficient B) 0<r<4 Since r is an integer, then the possible values that can take r are : 1,2 and 3 AS seen in Statement 1 , 2 and 3 are Pierpont prime but 1 is not prime , hence this statement is insufficent itself A+B) Now, Statements combined, we should have : 1<r<4 that give r the only two possible values : 2 and 3 and both of them are Pierpont prime as seen before Hence , the answer is Yes . Answer : C Hi, Can anyone explain, how can we have 1 as a solution using statement 2, since 2^0 X 3^0 + 1 = 2, so we can never reach 1. So only options left are 2 & 3 so ans should be statement B. As I see it, the only possibilities are: 2,3,4,5 So the correct answer is A. Can anyone confirm this?



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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14 Aug 2014, 15:55
PuneetSood wrote: Hi,
Can anyone explain, how can we have 1 as a solution using statement 2, since 2^0 X 3^0 + 1 = 2, so we can never reach 1. So only options left are 2 & 3 so ans should be statement B. Dear PuneetSood, I'm happy to respond. We CAN'T reach 1. That's the point. For any value of r, if r is prime and we can reach it using that formula, it's a Pierpont prime. But if r = 1, which is allowed by Statement #2, then we can't reach that form, and it's not prime anyway, so we get an answer of "no." If r = 1, is r a Pierpont prime? No. By convention, 1 is not a prime number at all. If r = 2, is r a Pierpont prime? Yes If r = 3, is r a Pierpont prime? Yes See: http://magoosh.com/gmat/2012/gmatmath ... menumber/Keep in mind the exact logical arrangement. We are not guaranteed that the r we pick will be a Pierpont prime. Instead, we are going to pick any possible r's in that range, and for each one, ask the question, "Is it a Pierpont prime?" Does this make sense? ronr34 wrote: As I see it, the only possibilities are: 2,3,4,5 So the correct answer is A. Can anyone confirm this? For statement #1, 1 < r < 5, the possible values of r are {2, 3, 4}. The value r = 5 is not a possibility. For these three values, we are asking the question: if r is this value, is it a Pierpoint prime? If r = 2, is r a Pierpont prime? Yes If r = 3, is r a Pierpont prime? Yes If r = 4, is r a Pierpont prime? NO! It's not a prime number at all. In order to be a Pierpoint prime, a number must (a) be a prime number, and (b) satisfy that equation. There are plenty of nonprime numbers that satisfy that equation  for starters, every power of 3 plus 1 (4, 10, 28, 82, 244, etc). If a number isn't prime, it can't be a Pierpont prime. Does this make sense? Mike
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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07 Nov 2015, 16:24
I had the same problem as PuneetSoodmikemcgarry, if 1 is prime or not is not the question here. The part I did not understood was: How can (2^k)(3^l) be 0? Since 2^0 = 1 and 3^0 = 1, I dont see how we can get to 1+1=1. Not being able to get to 1 would make B the right answer!



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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08 Nov 2015, 22:24
Mascarfi wrote: I had the same problem as PuneetSoodmikemcgarry, if 1 is prime or not is not the question here. The part I did not understood was: How can (2^k)(3^l) be 0? Since 2^0 = 1 and 3^0 = 1, I dont see how we can get to 1+1=1. Not being able to get to 1 would make B the right answer! Dear Mascarfi, I'm happy to respond. As I stated above, in order to be a Pierpont prime, an integer must satisfy two conditions: 1) it must be prime 2) it must satisfy the formula. You are perfectly correct that r = 1 does not satisfy the formula. There is absolutely no way to get it from the formula. I was focusing on the fact that 1 is not prime because that should be immediate. You should know without a moment's reflection that 1 absolutely is not a prime number, and therefore it cannot possibly be a Pierpont prime. You see, even checking whether you can generate it with the formula is more work that you should be doing. You should immediately recognize that 1 is not prime, and that should obviate any calculations. If you have to do even a single calculation, you have done too much to determine that 1 cannot possibly be a Pierpont prime. The fact that 1 cannot possibly be a Pierpont prime does NOT make Statement #2 sufficient by itself. You see, Statement #2 allows for three values. r = 1 > Is it a Pierpont prime? NO! It's not a prime at all. (Also, we can't get it from the formula) r = 2 > Is it a Pierpont prime? Yes. r = 3 > Is it a Pierpont prime? Yes. Different values of r give us different answers to the prompt, so Statement #2, alone and by itself, is not sufficient. We need to combine the statements so that only Pierpont primes, 2 and 3, are possible values of r. Does all this make sense? Mike
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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08 Nov 2015, 22:45



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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23 Sep 2017, 21:26
Hi mikemcgarry, Although I got the point of the question, I just have a small query that while solving the question I assumed that any number of the form given in the question is a Pierpont prime and not taking into consideration that the number should also be a prime number. How can we avoid such mistakes happening in future? Thanks in advance



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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24 Sep 2017, 17:36
devanshu92 wrote: Hi mikemcgarry, Although I got the point of the question, I just have a small query that while solving the question I assumed that any number of the form given in the question is a Pierpont prime and not taking into consideration that the number should also be a prime number. How can we avoid such mistakes happening in future? Thanks in advance Dear devanshu92, I'm happy to respond. Here is the first sentence. A Pierpont prime is any prime number p such that \(p=(2^k)(3^l)+1\) , where k and l are nonnegative integers. This states quite clearly that a Pierpont prime must be a prime number, not just any number that satisfies the formula. I have two responses to your question: 1) If a GMAT Quant prompt presents words and a formula, you NEVER can skip the words and jump to the formula. You have to read every word in the prompt with the same precise attention you give each number and formula. The exact wording in Quant prompts is crucial. 2) If you read that sentence and didn't understand its implications, it may be that you need to raise your reading level to excel on GMAT Quant. See: How to Improve Your GMAT Verbal ScoreDoes all this make sense? Mike
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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25 Sep 2017, 03:01
mikemcgarry wrote: devanshu92 wrote: Hi mikemcgarry, Although I got the point of the question, I just have a small query that while solving the question I assumed that any number of the form given in the question is a Pierpont prime and not taking into consideration that the number should also be a prime number. How can we avoid such mistakes happening in future? Thanks in advance Dear devanshu92, I'm happy to respond. Here is the first sentence. A Pierpont prime is any prime number p such that \(p=(2^k)(3^l)+1\) , where k and l are nonnegative integers. This states quite clearly that a Pierpont prime must be a prime number, not just any number that satisfies the formula. I have two responses to your question: 1) If a GMAT Quant prompt presents words and a formula, you NEVER can skip the words and jump to the formula. You have to read every word in the prompt with the same precise attention you give each number and formula. The exact wording in Quant prompts is crucial. 2) If you read that sentence and didn't understand its implications, it may be that you need to raise your reading level to excel on GMAT Quant. See: How to Improve Your GMAT Verbal ScoreDoes all this make sense? Mike Thanks Mike. It perfectly makes sense.



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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1 [#permalink]
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12 Feb 2018, 13:44
Hi All, We're told that a Pierpont prime is any PRIME number p such that p = (2^K)(3^L)+1, where K and L are nonnegative integers. We're told that R is an INTEGER. We're asked if R a Pierpont prime. This is a YES/NO question. 1) 1 < R < 5 From this Fact, R is limited to only 3 possibilities: 2, 3 and 4. We have to check to see if they fit the definition of a Pierpont prime... IF... R = 2, then K = 0 and L = 0 would give us R = (1)(1) + 1 = 2, so R IS a Pierpont Prime and the answer to the question is YES. R = 3, then K = 1 and L = 0 would give us R = (2)(1) + 1 = 3, so R IS a Pierpont Prime and the answer to the question is YES. R = 4, then the answer to the question is NO (since 4 is NOT a prime number) Fact 1 is INSUFFICIENT 2) 0 < R < 4 This Fact also limits R to only 3 possibilities: 1, 2 and 3. Our prior work (in Fact 1, above) will be useful here... IF.... R = 1, then the answer to the question is NO (since 1 is NOT a prime number) R = 2 or R = 3, then the answer to the question is YES (the work above proves this). Fact 2 is INSUFFICIENT Combined, we know... 1 < R < 5 0 < R < 4 R can ONLY be 2 or 3. Since both of those numbers lead to the same "YES" answer, the answer to the question is ALWAYS YES. Combined, SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1
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