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

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

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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

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

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New post Updated on: 07 Aug 2018, 05:34
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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

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Originally posted by EgmatQuantExpert on 10 Apr 2015, 07:38.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:34, edited 1 time in total.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 06:36
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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 :lol:

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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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 07:54
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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 :lol:

This is

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(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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 08:47
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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 09:11
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 ?
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 09:42
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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 11:14
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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 16:47
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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 09 Apr 2015, 19:58
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 :lol:

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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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 21 Aug 2015, 01:18
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
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 21 Aug 2015, 14:19
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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 06 Jan 2016, 21:13
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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 12 Jan 2016, 15:18
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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 03 Apr 2017, 01:23
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
Answer C.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 14 May 2017, 10:48
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?!
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 06 Aug 2018, 20:38
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?

Please help thanks
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime  [#permalink]

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New post 17 Aug 2018, 18:04
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.
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Re: The Prime Sum of an integer X greater than 1 is the sum of all prime &nbs [#permalink] 17 Aug 2018, 18:04
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