A Pierpont prime is any prime number p such that p = (2^k)(3^l) + 1

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?

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:
https://magoosh.com/gmat/2012/gmat-math- ... me-number/

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?


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 non-prime 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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I had the same problem as PuneetSood

mikemcgarry, 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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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 non-negative 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 Score

Does all this make sense?
Mike
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Thanks Mike. It perfectly makes sense.

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 non-negative 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

Hi,
I have a very basic doubt.
When we combine 1 < R < 5 & 0 < R < 4. Why are we taking the range from 1<r<4 and not 0<r<5 ?

Hi Karmesh,

When you have more than one 'restriction' on the range of a variable, you have to think in terms of the 'more limiting restriction(s)' - since that's the only way a variable will fit BOTH of the ranges.

For example, if we start with 1 < R < 5, then R COULD be 4. However, when we introduce the send restriction (re: 0 < R < 4), R can no longer equal 4.

The only way to account for the information in BOTH ranges is to take the most restrictive aspects of each - this gives us 1 < R < 4. With this range, ANY number you choose will automatically 'fit' both of the individual (less restrictive) ranges.

GMAT assassins aren't born, they're made,
Rich

Given: A Pierpont prime is any prime number p such that \(p =(2^k)(3^l)+1\), where k and l are non-negative integers.

Asked: If r is an integer, is r a Pierpont prime?

(1) 1 < r < 5
If r=2; r = 2^0*3^0 + 1 ; Pierpont prime
If r=3; r = 2^1*3^0 + 1; Pierpont prime
If r=4; r = 2^0*3^1 + 1; 4 is not prime; NOT Pierpont prime
NOT SUFFICIENT

(2) 0 < r < 4
If r=1; 2^k3^l = 0 ; 1 is NOT prime; NOT Pierpont prime
If r=2; r = 2^0*3^0 + 1 ; Pierpont prime
If r=3; r = 2^1*3^0 + 1; Pierpont prime
NOT SUFFICIENT

(1) + (2)
(1) 1 < r < 5
(2) 0 < r < 4
1<r<4
If r=2; r = 2^0*3^0 + 1 ; Pierpont prime
If r=3; r = 2^1*3^0 + 1; Pierpont prime
SUFFICIENT

IMO C

Hi in combining 1) & 2), why can't we add the both inequalities & get this:

0 < r < 4
1 < r < 5

1 <2r<9,
dividing by 2 : 0.5<r<4.5, so r = 1/2/3/4

I saw solutions adding 2 inequalities from both statements... please let me know why this is wrong...thanks

Hi serenayong,

When you have more than one 'restriction' on the range of a variable, you have to think in terms of the 'more limiting restriction(s)' - since that's the only way a variable will fit BOTH of the ranges.

For example, if we start with 1 < R < 5, then R COULD be 4. However, when we introduce the second restriction (re: 0 < R < 4), R can no longer equal 4.

The only way to account for the information in BOTH ranges is to take the most restrictive aspects of each - this gives us 1 < R < 4. With this range, ANY number you choose will automatically 'fit' both of the individual (less restrictive) ranges.

GMAT assassins aren't born, they're made,
Rich

I liked this question - for the reason I attempted this question in the mock, and I marked wrong! This question has a very nice trap!

Consider the definition of Pierpont prime. A prime number such that \(p =(2^k)(3^l)+1\)

Now, Statement 1 states that r={2,3,4,}
2 = (2^0)(3^0)+1 - fits in the expression
3 = (2^1)(3^0)+1 - fits in the expression
4 = (2^0)*3^1)+1 - fits in the expression

Statement 1 states that r={1,2,3,}
While 2 & 3 suite the expression, 1 cannot be written in this form, since Positive Integer raised to the power a non zero integer is always greater than zero. And 1 + a positive number is always greater than 1

So, Statement 1 is sufficient, while Statement 2 isn't right?

Wrong! Here is the trap! The definition of the Pierpont prime states that these numbers must be a prime number! So, 4 is definitely not a Pierpont prime. Thus, Statement 1 is insufficient.

Combining both: r = {2,3} - Both are Pierpont primes.

Therefore, answer is (C)

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