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# If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,

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Joined: 19 May 2019
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If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,  [#permalink]

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06 Aug 2019, 18:40
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55% (hard)

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46% (01:16) correct 54% (01:27) wrong based on 26 sessions

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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

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?
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Re: If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,  [#permalink]

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06 Aug 2019, 19:18
1
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.
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Re: If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,  [#permalink]

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07 Aug 2019, 03:13
1
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!
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Re: If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,   [#permalink] 07 Aug 2019, 03:13
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# If k is a positive integer, is k^(1/2) an integer? 1) k = m^2,

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