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X is a number which on squaring produces Y. If Y has 3 factors, how

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X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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New post Updated on: 13 Aug 2018, 02:20
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Question Stats:

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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

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Question 6 of The e-GMAT Number Properties Marathon




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Question 7 of the Marathon


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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.
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Re: X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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New post 27 Feb 2018, 11:13
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EgmatQuantExpert wrote:

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


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
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Re: X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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New post 28 Feb 2018, 00:17
1

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:
       \((a+1) * (b+1) = 3*1\)
       \((a+1) = 3\); \((b+1) =1\)
    => \(a=2, b=0\)
    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, \(Y= X^2 = P1^2\)
    =>\(X^2 = P1^2\)
    => \(X=P1\) (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.
Therefore, \(8\) such numbers lie among the first \(20\) natural numbers.
Correct Answer: Option (E)
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Re: X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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Re: X is a number which on squaring produces Y. If Y has 3 factors, how   [#permalink] 07 Sep 2019, 00:01
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