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05 Sep 2022, 04:37
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Difficulty:
55%
(hard)
Question Stats:
56% (01:31) correct
44%
(01:50)
wrong
based on 140
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Is the positive square root of x an integer?
(1) \(x = n^3\) and n is a perfect square.
(2) \(x\) has exactly seven factors.
(1) \(x = n^3\) and n is a perfect square.
(2) \(x\) has exactly seven factors.
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05 Sep 2022, 06:45
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(1) x=n^3 and n is a perfect square.
As n is a perfect square, let \(n = m^2\)
Which means that \(x = (m^2)^3\)
\(x = m^6\)
taking the square root of both sides will give: \(\sqrt{x} = m^3\)
As we know that \(m\) is an integer, \(m^3\) will definitely be an integer and therefore the positive square root of \(x\) will also be an integer.
Sufficient
(2) x has exactly seven factors.
As 7 is a prime number, the only possible way a number will have 7 factors is if the number has a single prime factor with a power of 6. This means that the square root of that number will be its prime factor to the power of 3, which will definitely be an integer.
Sufficient
Answer D
As n is a perfect square, let \(n = m^2\)
Which means that \(x = (m^2)^3\)
\(x = m^6\)
taking the square root of both sides will give: \(\sqrt{x} = m^3\)
As we know that \(m\) is an integer, \(m^3\) will definitely be an integer and therefore the positive square root of \(x\) will also be an integer.
Sufficient
(2) x has exactly seven factors.
As 7 is a prime number, the only possible way a number will have 7 factors is if the number has a single prime factor with a power of 6. This means that the square root of that number will be its prime factor to the power of 3, which will definitely be an integer.
Sufficient
Answer D
05 Sep 2022, 07:00
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Is the positive square root of x an integer?
Two properties of Perfect Square which will be handy in solving the question
- Perfect Square is true for integers only
- The number of distinct factors of a Perfect Square is Always Odd
(1) \(x=n^3\) and n is a perfect square.
Since n is a perfect square, say n = \(a^2\), where a is an integer
then, x = \(n^3\) = \((a^2)^3\) = \(a^6\)
and \(sqrt x = sqrt(a^6) = a^3 \)-> which is an integer. Hence st1 is Sufficient.
(2) x has exactly seven factors.
since x has odd number of distinct factors (7) therefore, x is a Perfect Square
\(x = b^2\), where b is an integer
and \(sqrt x = sqrt(b^2) = b\) -> which is an integer. Hence st2 is Sufficient.
Answer: D
Two properties of Perfect Square which will be handy in solving the question
- Perfect Square is true for integers only
- The number of distinct factors of a Perfect Square is Always Odd
(1) \(x=n^3\) and n is a perfect square.
Since n is a perfect square, say n = \(a^2\), where a is an integer
then, x = \(n^3\) = \((a^2)^3\) = \(a^6\)
and \(sqrt x = sqrt(a^6) = a^3 \)-> which is an integer. Hence st1 is Sufficient.
(2) x has exactly seven factors.
since x has odd number of distinct factors (7) therefore, x is a Perfect Square
\(x = b^2\), where b is an integer
and \(sqrt x = sqrt(b^2) = b\) -> which is an integer. Hence st2 is Sufficient.
Answer: D
