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Updated on: 13 Aug 2018, 02:20
Originally posted by EgmatQuantExpert on 27 Feb 2018, 10:40.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:20, edited 3 times in total.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:20, edited 3 times in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (01:38) correct
37%
(01:50)
wrong
based on 799
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e-GMAT Question:
X is a number which on squaring produces Y. If Y has 3 factors, how many such X are present in the first 20 natural numbers?
- A) 2
B) 4
C) 5
D) 7
E) 8
This is
Question 6 of The e-GMAT Number Properties Marathon
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27 Feb 2018, 11:13
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EgmatQuantExpert
given \(x^2=y\)
\(y\) has only 3 factors implies that \(y\) is a square of a prime number, Hence \(y=p^2\), where \(p\) is some prime number
Hence \(x^2=p^2 => x\) is a prime number.
Prime numbers less than 20 are: 2,3,5,7,11,13,17,19. total \(8\)
Option E
28 Feb 2018, 00:17
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Solution:
The question specifies that the square of \(X\): \(Y\), has \(3\) factors.
- o So, \(1\) and \(Y\) must be the factors of \(Y\).
- Apart from these two, \(Y\) has one more factor.
- o Per the conceptual knowledge of Prime factorization, a number can be represented in the form of: \(P1^a * P2^b * P3^c…Pn^z\), where \(P1, P2,P3\).. are the prime factors of the number \(X\) and \(a, b, c\).. are the powers of the prime factors.
- o Total factors of this number is written as: \((a+1) * (b+1)\)…
- Since total factors = \(3\), let us see how can \(3\) be formed using the above formula:
- • \(3 = 3*1\)
- This is the only possible combination.
- o Now, putting this in the formula of total factors we get:
- \((a+1) * (b+1) = 3*1\)
\((a+1) = 3\); \((b+1) =1\)
- o Thus, the number \(Y\) can be written as \(P1^2\) \(* P2^0\), where \(P1\)and \(P2\) are the prime factors of the number \(Y\).
- However, \(P2^0\) will simplify to \(1\), which will further reduce the overall equation to: \(Y = P1^2\).
- • Since, we also know that \(Y=X^2\)
• Therefore, \(Y= X^2 = P1^2\)
=> \(X=P1\) (Cannot take the negative value since prime numbers are positive numbers)
- This means \(X\) is nothing but a prime number.
- • This was the observation that was needed to solve this question.
• Now, the entire sum reduces to a simple problem of prime numbers, i.e.: How many prime numbers exist between \(1\) and \(20\), inclusive?
- o Prime numbers between \(1\) and \(20\), inclusive, are: \(2,3,5,7,11,13,17,19\): \(8\) primes.
Correct Answer: Option (E)
