The Prime Sum of an integer X greater than 1 is the sum of all prime

Question

The Prime Sum of an integer X greater than 1 is the sum of all prime factors of X including repetitions. A positive integer n can be expressed as a product of two natural numbers at least one of which is even. What is the Prime Sum of n?

(1) n has only 3 prime factors and the smallest prime factor of n is raised to a power equal to the next biggest prime factor of n.

(2) The prime factors of n are consecutive and the total number of divisors of n is 16

We will provide the OA in some time. Till then Happy Solving :lol:

This is

Ques 6 of The E-GMAT Number Properties Knockout

Register for our Free Session on Number Properties this Saturday to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts!

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-NP-e1533292073470.jpg

EgmatQuantExpert · Accepted Answer

Detailed Solution

Step-I: Given Info

We are given that a positive integer  n  can be expressed as a product of two natural numbers at least one of which is even. We are asked to find the Prime Sum of  n .

Step-II: Interpreting the Question Statement

For finding the prime sum of  n , we need to know all the prime factors of  n  along with the no. of times they are repeated in the prime factorization of  n .

For example if  n= P_{1}^a*P_{2}^b , we need to find the values of  P_{1} ,  P_{2}  along with the values of  a  and  b  to calculate the prime sum of the number.

We are told that  n  has an even number as it factor, which implies that  n  is even. Hence, we can say with certainty that 2 is one of the prime factors of  n . Now, we need to find the other prime factors of  n  along with the no. of times they are repeated.

Step-III: Statement I

We know from statement-I that  n  has 3 prime factors. So, we can write  n  as:

n= P_{1}^a* P_{2}^b * P_{3}^c where  P_{1} ,  P_{2}  and  P_{3}  are prime numbers.

The statement also tells us that the smallest prime factor of  n  which is 2 (as 2 is the prime factor of  n  and there is no smaller prime number than 2) is raised to a power equal to the next biggest prime factor of  n . So  n  can be rephrased as:

n= 2^{P_{2}} * P_{2}^b * P_{3}^c

But, we don’t have any information about the values of other prime factors of  n  as well as the no. of times they are repeated.
So, Statement-I is insufficient to answer the question.

Step-IV: Statement II

Statement-II tells us that the total number of divisors of  n  is 16, so n can be written as:

  n= P_{1}* P_{2} * P_{3} * P_{4}  (as 16 = 2*2*2*2) or 
  n= P_{1}^a* P_{2}^b * P_{3}^c  (as 16 = 4*2*2) where either of  a ,  b  or  c  can be equal to 3 and rest as 1 or
  n= P_{1}^3* P_{2}^3  (as 16 = 4*4)

We know from the question statement that  P_{1}  = 2. Since the statement tells us that the prime factors of  n  are consecutive, the prime factors may be:

 2,3,5,7 or 
 2,3,5 or
 2,3

We can only know from case III the required values of the prime factors and the number of times they are repeated. The other two cases do not provide us with a unique answer.
Thus Statement-II is insufficient to answer the question.

Step-V: Combining Statements I & II

If we combine statements-I & II, we can notice that  n  has 3 prime factors, so  n  can be written as:

n= 2^{P_{2}}* P_{2}^b * P_{3}^c . Since the prime factors of n are consecutive that would imply  P_{2}  = 3 and  P_{3}  = 5.

Also, St-I tells us that 2 is raised to the power equal to the next biggest prime factor of  n  i.e. 3, so, n can be written as:

n= 2^3* 3^1 *5^1

Now, we know the prime factors of  n  and the no. of times they are repeated, we can definitely find the prime sum of  n .

Thus combination of St-I & II is sufficient to answer the question.
Hence the correct answer is  Option C

Key Takeaways

1. 2 is the smallest prime number
2. If the total number of factors of a number is given to be T, evaluate the possible ways in which T can be expressed as a product of 2 or more numbers

Amit0507 - Kudos for pointing out the difference between consecutive prime numbers and consecutive natural numbers. This was one of the key concepts being checked here.

Regards
Harsh

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

Bunuel · Answer

Question on the "prime sum" from OG to practice: the-prime-sum-of-an-integer-n-greater-than-1-is-the-sum-of-167088.html

Lucky2783 · Answer

(1) n has only 3 prime factors and the smallest prime factor of n is raised to a power equal to the next biggest prime factor of n.

Given N is EVEN .

so one prime factor is 2 and this is also the smallest . 2 is smallest prime.

next biggest prime can be 3,5,7,11,..... such that number will be

N=2*  5^2 , 3^2 , 7^2

insufficient.

(2) The prime factors of n are consecutive and the total number of divisors of n is 16
 
 if N= 2^3 * 3^3  #factors =16 prime sum= 2*3+3*3= 15 
 
 if N= 2^1 * 3^7  #factors = 16 primesum = 2+ 3*7 = 23
 
 Prime sum will be different for two numbers. 
 
 insufficient.

combining together we know there are 3 prime factor and prime factors are consecutive and that smallest prime is raised to power of next biggest prime . 
 
 only 2 and 3 are prime factors which are consecutive . 
 
 so number should be  2*3^2  
 3 prime factors are 2,3,3 
 smallest prime is raised to power of next biggest  2, 3^2 
 prime factors are consecutive .
 but in this case # of factors are not 16.

Answer should be E.

chandru42 · Answer

I ll go with C

Given that n is even so one of the prime factors is 2.

From 1

n Can be 2^3*3 or 2^5*5 ( Prime Sum Different) Insufficient

From 2

n can be 2^3*3^3 or 2*3^3*5 ( Prime Sum Different) Insufficient

From 1 and 2

The prime factors are 2, 3, 5

Exponent of 2 is 3
Therefore product of exponent of 3 and 5 is 4 which is possible only when 3*5

So Prime Sum = 2^3+3+5

Sufficient

Lucky2783 · Answer

The prime factors of n are consecutive so i am not sure how 'The prime factors are 2, 3, 5' .

am i missing something ?

chandru42 · Answer

From 1, n has only 3 prime factors. From 2 the prime factors are consecutive. 
n is even so it must have 2 as its prime factor.
So the prime factors can be 2,3,5

Lucky2783 · Answer

I think i misunderstood stmt2 And thought that only 2 and 3 are consecutive and that all consecutive numbers has a common difference.

(2) The prime factors of n are consecutive and the total number of divisors of n is 16

Amit0507 · Answer

Consecutive prime factors are different from consecutive numbers. 5,7,11 are consecutive primes but not conse. numbers. The answer should be C. 2^3.3.5

tronghieu1987 · Answer

I go to C
n is expressed as a product of 2 numbers, at least one of which is even -> n has 2 as one of it's prime factor.

(1) "n" must equal: 2^x * x^a * y^b (where x and y are prime numbers others than 2). No other clues about x and y -> NOT SUFFICIENT

(2) The total number of divisors is 16 (2*2*2*2 or 2*2*4 or 2*4*2 or 4*2*2), so the powers of prime factors of n can be ( 1, 1, 1, 1 or 1, 1, 3 or 1, 3, 1 or 3, 1, 1). NOT SUFFICIENT

(1) and (2) together: "n" must equal: 2^3 * 3^a * 5^b and the powers (3, a, b) must be one of 4 scenarios above -> must be 3,1,1. 
So n = 2^3 * 3 * 5. SUFFICIENT -> C is the correct answer.

PerptualKnite56 · Answer

The Prime Sum of an integer X greater than 1 is the sum of all prime factors of X including repetitions. A positive integer n can be expressed as a product of two natural numbers at least one of which is even. What is the Prime Sum of n?

(1) n has only 3 prime factors and the smallest prime factor of n is raised to a power equal to the next biggest prime factor of n.

(2) The prime factors of n are consecutive and the total number of divisors of n is 16

1) n= 2^3*3 ,2*5,5 so A insuff
2) Given that "prime factors of n are consecutive". Only set of prime factors that are consecutive are 2 & 3. If we consider 5 as one factor of n then its are not consecutive. 
Suppose if it is mentioned as "consecutive prime numbers" then we can consider 5 as one factor.(Pl correct me if I'm wrong.)
So n should have only 2 prime factors 2 & 3
16 can be expressed as 4*4 = (3+1)(3+1) hence n = 2^3*3*3 
We can find the answer, hence B is suff
Ans: B

MasterGMan · Answer

statement 2 says number "n" has 16 divisors, these 16 divisors will include negative divisors as well (assuming word "divisor" means the same as "factors") So, no. of +ve divisors of "n" will be 8. And 2^3 X 3^a X 5^b ==> aXb = 2. And this is just not possible here as both a and b must atleast be equal to 1 for 3 prime factors to exist
C would have been correct if statement 2 said total no of +ve divisors are 16.

jwam2 · Answer

Quick question, for statement 2, can there be another option for the two number case? (i.e. x^1*y^7)

Also, it seems for this question, we needed to find the solution 2^3*3*5 to answer this correctly. I haven't seen that as much for DS, is this the only way to find the answer or could we have stopped at some point earlier to understand we need both statements to have sufficient information.

PerptualKnite56 · Answer

The second option clearly states , which means the prime factors  happen to be consecutive  and only prime numbers that are consecutive are 2 & 3.
The statement 2 does not state  consecutive prime factors , in which case we can consider the combination 2,3,5 as a possibility but it is not the case.
So option B should be correct.

Alexey1989x · Answer

Stem tells us that 1 of the prime factors of n is even = 2.
(1)n=2*b*c,
b and c can be any prime numbers, hence  Insuff. 
(2) 
1 case: n = 2^3*3^3
number of factors=(3+1)*(3+1)=16
2 case: n=2^1*3^7
number of factors=(1+1)*(7+1)=16
(1)+(2):
Combining all constraints we have, that N consists of 3 consecutive prime factors, one of which is 2 which, in turn, is raised to a power equal to next highest prime factor and total number of factors is 16.
n=2^3*3*5
number of factors=(3+1)*(1+1)*(1+1)=16
Length=6+3+5=14
 Answer C.

anishkhadgi · Answer

Statement-II tells us that the total number of divisors of nn is 16, so n can be written as:

 n=P1∗P2∗P3∗P4n=P1∗P2∗P3∗P4 (as 16 = 2*2*2*2) or
 n=Pa1∗Pb2∗Pc3n=P1a∗P2b∗P3c (as 16 = 4*2*2) where either of aa, bb or cc can be equal to 3 and rest as 1 or
 n=P31∗P32n=P13∗P23 (as 16 = 4*4)

in above there can be 8*2 options as well which makes option 3 also indeterministic.is that correct?!

zanaik89 · Answer

Hi Harsh,

Statement-II tells us that the total number of divisors of n is 16, so n can be written as:

 n=P1∗P2∗P3∗P4(as 16 = 2*2*2*2) or 
 n=Pa1∗Pb2∗Pc3n(as 16 = 4*2*2) where either of aa, bb or cc can be equal to 3 and rest as 1 or
 n=P31∗P32 (as 16 = 4*4)

You r saying that the total no. of divisors of n-16 which means N is divisible by 16 divisors right?

Please help thanks

shmba · Answer

i think the key is to understand what 'consecutive prime factor' is. it should refer to 2,3,5 as such instead of 2,3 only.

zhikyang · Answer

hi Folks,

Can someone clarify in statement (1) what does it mean by:the smallest prime factor of n is raised to a power equal to the next biggest prime factor of n?

I thought the smaller prime factor of n = 2, so it says 2^x = next biggest prime factor?

Thanks

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

The Prime Sum of an integer X greater than 1 is the sum of all prime

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

The Prime Sum of an integer X greater than 1 is the sum of all prime

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards