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06 Aug 2019, 18:40
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A
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Difficulty:
35%
(medium)
Question Stats:
61% (01:14) correct
39%
(01:28)
wrong
based on 92
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If k is a positive integer, is k^(1/2) an integer?
1) k = m^2, where m is an integer
2) k^(1/2) = n, where n^2 is an integer
Source: Jeff Sackmann - Number Properties Challenge
1) k = m^2, where m is an integer
2) k^(1/2) = n, where n^2 is an integer
Source: Jeff Sackmann - Number Properties Challenge
I'm getting a different answer than the official answer. I understand statement 1 and why it is sufficient, but when I look at statement 2, I'm reading it as n^2 = intg. Therefore, n^2 must be perfect square. E.g. n^2 = 36, and if we take the square root of both sides, we get n = 6. This means that \sqrt{k} = integer.
This leads me to an answer of D, but the official answer says A. Why is it incorrect to view n^2 = perfect square? Can someone please clarify?
This leads me to an answer of D, but the official answer says A. Why is it incorrect to view n^2 = perfect square? Can someone please clarify?
06 Aug 2019, 19:18
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From 1:
\(k = m^2\). Then \(\sqrt{k}\) = m. As m is an integer. \(\sqrt{k}\) is also an integer.
Sufficient.
From 2:
\(\sqrt{k} = n\). Where \(n^2\) is an integer.
If n = 2. Then yes
If n = \(\sqrt{2}\), then NO.
Insufficient.
A is the answer.
\(k = m^2\). Then \(\sqrt{k}\) = m. As m is an integer. \(\sqrt{k}\) is also an integer.
Sufficient.
From 2:
\(\sqrt{k} = n\). Where \(n^2\) is an integer.
If n = 2. Then yes
If n = \(\sqrt{2}\), then NO.
Insufficient.
A is the answer.
07 Aug 2019, 03:13
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When you are attempting a DS question, always look for a way of rephrasing the question without changing its meaning.
For instance, in this question, we are supposed to find out if \(k^\frac{1}{2}\) is an integer. In other words, we are trying to find out if ‘k’ is a perfect square. Because only when ‘k’ is a perfect square will \(k^\frac{1}{2}\) be an integer.
From statement I, we can easily infer that k is a perfect square. This is because it is equated to a perfect square, \(m^2\).
So, now, \(k^\frac{1}{2}\) = \((m^2)^\frac{1}{2}\) = m, which is an integer.
Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be eliminated.
From statement II, although we know that \(n^2\) is an integer, we do not know anything about n.
If \(k^\frac{1}{2}\) = n, squaring both sides will give us k = \(n^2\), which is given as an integer. Although we know that k is an integer, it does not imply that \(k^\frac{1}{2}\) will always be , an integer.
For example, if k = \(n^2\) = 2, then \(k^\frac{1}{2}\) = √2, which is not an integer.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
Hope this helps!
For instance, in this question, we are supposed to find out if \(k^\frac{1}{2}\) is an integer. In other words, we are trying to find out if ‘k’ is a perfect square. Because only when ‘k’ is a perfect square will \(k^\frac{1}{2}\) be an integer.
From statement I, we can easily infer that k is a perfect square. This is because it is equated to a perfect square, \(m^2\).
So, now, \(k^\frac{1}{2}\) = \((m^2)^\frac{1}{2}\) = m, which is an integer.
Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be eliminated.
From statement II, although we know that \(n^2\) is an integer, we do not know anything about n.
If \(k^\frac{1}{2}\) = n, squaring both sides will give us k = \(n^2\), which is given as an integer. Although we know that k is an integer, it does not imply that \(k^\frac{1}{2}\) will always be , an integer.
For example, if k = \(n^2\) = 2, then \(k^\frac{1}{2}\) = √2, which is not an integer.
Statement II alone is insufficient. Answer option D can be eliminated.
The correct answer option is A.
Hope this helps!
