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16 Sep 2014, 01:44
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If \([x]\) denotes the greatest integer less than or equal to \(x\) for any number \(x\), is \([a] + [b] = 1\)?
(1) \(ab = 2\).
(2) \(0 < a < b < 2\).
(1) \(ab = 2\).
(2) \(0 < a < b < 2\).
16 Sep 2014, 01:44
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Official Solution:
If \([x]\) denotes the greatest integer less than or equal to \(x\) for any number \(x\), is \([a] + [b] = 1\)?
We are given that the function [ ] rounds DOWN a number to the nearest integer. For example, \([1.5]=1\), \([2]=2\), \([-1.5]=-2\), ...
(1) \(ab = 2\): The above implies that \(a\) and \(b\) are of the same sign. Consider two cases:
If \(a\) and \(b\) are both negative, then \([a]+[b]\) cannot be positive. Therefore, in this case, \([a]+[b]\neq 1\). Hence, if both \(a\) and \(b\) are negative, \([a] + [b] \neq 1\) and we have a NO answer to the question. To see why \([a]+[b]\) cannot be positive for negative \(a\) and \(b\), note that the maximum value of \([a] + [b]\) will occur when \(-1 \leq a < 0\) and \(-1 \leq b < 0\). In this case, \([a] + [b] = -1 + (-1) = -2\).
If \(a\) and \(b\) are both positive, then for \([a]+[b]=1\) to hold true, we must have \([a]=0\) and \([b]=1\) (or vice versa). This means \(0 \leq a \lt 1\) and \(1 \leq b \lt 2\) (or vise-versa). However, in this case, \(ab\) cannot equal 2. Therefore, if both \(a\) and \(b\) are positive and \(ab = 2\), \([a] + [b] \neq 1\), and we have a NO answer to the question.
Both cases give the same NO answer to the question: \([a] + [b] \neq 1\). Hence, (1) is sufficient.
(2) \(0 \lt a \lt b \lt 2\).
If \(a=\frac{1}{2}\) and \(b=1\), then \([a] + [b] = 0 + 1 = 1\). But if \(a=\frac{1}{4}\) and \(b=\frac{1}{2}\), then \([a] + [b] = 0 + 0 = 0\). Therefore, (2) is not sufficient to answer the question.
Answer: A
If \([x]\) denotes the greatest integer less than or equal to \(x\) for any number \(x\), is \([a] + [b] = 1\)?
We are given that the function [ ] rounds DOWN a number to the nearest integer. For example, \([1.5]=1\), \([2]=2\), \([-1.5]=-2\), ...
(1) \(ab = 2\): The above implies that \(a\) and \(b\) are of the same sign. Consider two cases:
If \(a\) and \(b\) are both negative, then \([a]+[b]\) cannot be positive. Therefore, in this case, \([a]+[b]\neq 1\). Hence, if both \(a\) and \(b\) are negative, \([a] + [b] \neq 1\) and we have a NO answer to the question. To see why \([a]+[b]\) cannot be positive for negative \(a\) and \(b\), note that the maximum value of \([a] + [b]\) will occur when \(-1 \leq a < 0\) and \(-1 \leq b < 0\). In this case, \([a] + [b] = -1 + (-1) = -2\).
If \(a\) and \(b\) are both positive, then for \([a]+[b]=1\) to hold true, we must have \([a]=0\) and \([b]=1\) (or vice versa). This means \(0 \leq a \lt 1\) and \(1 \leq b \lt 2\) (or vise-versa). However, in this case, \(ab\) cannot equal 2. Therefore, if both \(a\) and \(b\) are positive and \(ab = 2\), \([a] + [b] \neq 1\), and we have a NO answer to the question.
Both cases give the same NO answer to the question: \([a] + [b] \neq 1\). Hence, (1) is sufficient.
(2) \(0 \lt a \lt b \lt 2\).
If \(a=\frac{1}{2}\) and \(b=1\), then \([a] + [b] = 0 + 1 = 1\). But if \(a=\frac{1}{4}\) and \(b=\frac{1}{2}\), then \([a] + [b] = 0 + 0 = 0\). Therefore, (2) is not sufficient to answer the question.
Answer: A
JackSparr0w
26 Nov 2014, 09:04
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For stmt 1, wouldn't the max value of [a]+[b] be -3? because (-1)(-2)=2, and (-1)+(-2)=-3?
Or am I missing some factors?
Thanks
Or am I missing some factors?
Thanks
