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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3078
X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 54% (01:36) correct 46% (01:41) wrong based on 217 sessions

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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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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.
Retired Moderator D
Joined: 25 Feb 2013
Posts: 1177
Location: India
GPA: 3.82
Re: X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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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
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3078
Re: X is a number which on squaring produces Y. If Y has 3 factors, how  [#permalink]

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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:
 $$(a+1) * (b+1) = 3*1$$
 $$(a+1) = 3$$; $$(b+1) =1$$
=> $$a=2, b=0$$
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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