GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Dec 2018, 09:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### 10 Keys to nail DS and CR questions

December 17, 2018

December 17, 2018

06:00 PM PST

07:00 PM PST

Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.
• ### R1 Admission Decisions: Estimated Decision Timelines and Chat Links for Major BSchools

December 17, 2018

December 17, 2018

10:00 PM PST

11:00 PM PST

From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.

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

Author Message
TAGS:

### Hide Tags

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2324
The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

Updated on: 13 Aug 2018, 02:10
5
41
00:00

Difficulty:

95% (hard)

Question Stats:

55% (02:34) correct 45% (02:25) wrong based on 705 sessions

### HideShow timer Statistics

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

This is

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

_________________

Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Originally posted by EgmatQuantExpert on 09 Apr 2015, 03:52.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:10, edited 5 times in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2324
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

Updated on: 07 Aug 2018, 04:34
4
4
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

_________________

Number Properties | Algebra |Quant Workshop

Success Stories
Guillermo's Success Story | Carrie's Success Story

Ace GMAT quant
Articles and Question to reach Q51 | Question of the week

Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities

Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Originally posted by EgmatQuantExpert on 10 Apr 2015, 06:38.
Last edited by EgmatQuantExpert on 07 Aug 2018, 04:34, edited 1 time in total.
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 51263
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 05:36
1
1
EgmatQuantExpert wrote:
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

This is

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!

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
_________________
Director
Joined: 07 Aug 2011
Posts: 536
GMAT 1: 630 Q49 V27
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 06:54
1
1
EgmatQuantExpert wrote:
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

This is

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!

(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.

_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

Manager
Joined: 21 May 2009
Posts: 93
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 07:47
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
Director
Joined: 07 Aug 2011
Posts: 536
GMAT 1: 630 Q49 V27
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 08:11
chandru42 wrote:
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

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

am i missing something ?
_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

Manager
Joined: 21 May 2009
Posts: 93
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 08:42
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
Director
Joined: 07 Aug 2011
Posts: 536
GMAT 1: 630 Q49 V27
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 10:14
chandru42 wrote:
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

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
_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

Manager
Joined: 26 May 2013
Posts: 52
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 15:47
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

Lucky2783 wrote:
chandru42 wrote:
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

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
Current Student
Joined: 29 Apr 2014
Posts: 120
Location: Viet Nam
Concentration: Finance, Technology
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q51 V27
GMAT 3: 680 Q50 V31
GMAT 4: 710 Q50 V35
GMAT 5: 760 Q50 V42
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

09 Apr 2015, 18:58
EgmatQuantExpert wrote:
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

This is

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!

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.
Intern
Joined: 07 Jun 2015
Posts: 7
WE: Information Technology (Computer Software)
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

21 Aug 2015, 00:18
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
Current Student
Joined: 26 Jul 2015
Posts: 19
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

21 Aug 2015, 13:19
EgmatQuantExpert wrote:
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

Regards
Harsh

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.
Intern
Joined: 08 Dec 2015
Posts: 23
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

06 Jan 2016, 20:13
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.
Intern
Joined: 07 Jun 2015
Posts: 7
WE: Information Technology (Computer Software)
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

12 Jan 2016, 14:18
The second option clearly states
Quote:
(2) The prime factors of n are consecutive and the total number of divisors of n is 16
, 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.
Manager
Joined: 05 Dec 2016
Posts: 244
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

03 Apr 2017, 00:23
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
Intern
Joined: 07 May 2017
Posts: 1
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

14 May 2017, 09:48
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?!
Manager
Joined: 19 Aug 2016
Posts: 84
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

06 Aug 2018, 19:38
EgmatQuantExpert wrote:
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

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?

Intern
Joined: 16 Aug 2018
Posts: 28
Concentration: General Management, Strategy
Schools: Guanghua"21 (A)
GMAT 1: 700 Q49 V36
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

17 Aug 2018, 17:04
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.
Intern
Joined: 21 Oct 2018
Posts: 5
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

### Show Tags

10 Nov 2018, 14:54
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
Re: The Prime Sum of an integer X greater than 1 is the sum of all prime &nbs [#permalink] 10 Nov 2018, 14:54
Display posts from previous: Sort by