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X is a number which on squaring produces Y. If Y has 3 factors, how
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Updated on: 13 Aug 2018, 01:20
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Re: X is a number which on squaring produces Y. If Y has 3 factors, how
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27 Feb 2018, 10:13
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? 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
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27 Feb 2018, 23:17
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: 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 &nbs
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