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13 Feb 2022, 05:52
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
56% (01:31) correct
44%
(00:55)
wrong
based on 41
sessions
History
Date
Time
Result
Not Attempted Yet
Is integer \(x\) a perfect cube?
1) The tens digit of \(x\) is \(2\).
2) The units digit of \(x\) is \(0\).
1) The tens digit of \(x\) is \(2\).
2) The units digit of \(x\) is \(0\).
13 Feb 2022, 06:48
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Statement 1:
If \(x=27\), then the answer to the question is yes.
If \(x=28\), then the answer to the question is no.
Not Sufficient
Statement 2:
If \(x=1000\), then the answer to the question is yes.
If \(x=100\), then the answer to the question is no.
Not Sufficient
Combined:
In order for \(x\) to be a perfect cube with a units digit of \(0\), then \(x\) must be the cube of a number with a units digit of \(0\). But think about what happens when you cube a number with a units digit of \(0\):
\(10^3=1,000\)
\(20^3=8,000\)
\(100^3=1,000,000\)
\(200^3=8,000,000\)
In fact, we can rewrite any integer \(k\) that ends in \(n\) zeros as \(k=a\times10^n\), where \(a\) is some integer.
And when we cube \(k\), we get \(k^3=(a\times10^n)^3=a^3\times10^{3n}\)
The number of \(0\)s at the end of a perfect cube must be a multiple of \(3\).
Since we know that the last two digits of \(x\) are \(20\), we know that \(x\) ends in exactly one zero, and thus that \(x\) cannot be a perfect cube. Sufficient
If \(x=27\), then the answer to the question is yes.
If \(x=28\), then the answer to the question is no.
Not Sufficient
Statement 2:
If \(x=1000\), then the answer to the question is yes.
If \(x=100\), then the answer to the question is no.
Not Sufficient
Combined:
In order for \(x\) to be a perfect cube with a units digit of \(0\), then \(x\) must be the cube of a number with a units digit of \(0\). But think about what happens when you cube a number with a units digit of \(0\):
\(10^3=1,000\)
\(20^3=8,000\)
\(100^3=1,000,000\)
\(200^3=8,000,000\)
In fact, we can rewrite any integer \(k\) that ends in \(n\) zeros as \(k=a\times10^n\), where \(a\) is some integer.
And when we cube \(k\), we get \(k^3=(a\times10^n)^3=a^3\times10^{3n}\)
The number of \(0\)s at the end of a perfect cube must be a multiple of \(3\).
Since we know that the last two digits of \(x\) are \(20\), we know that \(x\) ends in exactly one zero, and thus that \(x\) cannot be a perfect cube. Sufficient
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