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Updated on: 13 Aug 2018, 03:10
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.
Last edited by EgmatQuantExpert on 13 Aug 2018, 03:10, edited 5 times in total.
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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
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 \(n\).
(2) The prime factors of \(n\) are consecutive and the total number of divisors of \(n\) is 16
This is
Ques 6 of The E-GMAT Number Properties Knockout
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Updated on: 07 Aug 2018, 05:34
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.
Last edited by EgmatQuantExpert on 07 Aug 2018, 05:34, edited 1 time in total.
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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
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
General Discussion
09 Apr 2015, 06:36
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EgmatQuantExpert
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
