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08 Nov 2021, 09:45
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Is a two digit positive integer \(mn\), where \(m\) is the tens digit and \(n\) is the units digits, odd?
(1) The least common multiple of \(m\) and \(n\) is even
(2) \(m\) is an odd number
(1) The least common multiple of \(m\) and \(n\) is even
(2) \(m\) is an odd number
08 Nov 2021, 09:45
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Official Solution:
Is a two digit positive integer \(mn\), where \(m\) is the tens digit and \(n\) is the units digits, odd?
Notice that the question basically asks whether \(n\) is odd (the units digit of an integer determines eve/odd nature of that integer).
(1) The least common multiple of \(m\) and \(n\) is even
The above means that at least one of \(m\) and \(n\) is even (else how/why the LCM would be even?). Thus, either both \(m\) and \(n\) are even OR one is even and another is odd. So, \(n\) can be even as well as odd. Not sufficient.
(2) \(m\) is an odd number
Clearly insufficient.
(1)+(2) From (2) we know that \(m\) is NOT even and from (1) we know that at least one of \(m\) and \(n\) must be even, thus \(n\) is even. Sufficient.
Answer: C
Is a two digit positive integer \(mn\), where \(m\) is the tens digit and \(n\) is the units digits, odd?
Notice that the question basically asks whether \(n\) is odd (the units digit of an integer determines eve/odd nature of that integer).
(1) The least common multiple of \(m\) and \(n\) is even
The above means that at least one of \(m\) and \(n\) is even (else how/why the LCM would be even?). Thus, either both \(m\) and \(n\) are even OR one is even and another is odd. So, \(n\) can be even as well as odd. Not sufficient.
(2) \(m\) is an odd number
Clearly insufficient.
(1)+(2) From (2) we know that \(m\) is NOT even and from (1) we know that at least one of \(m\) and \(n\) must be even, thus \(n\) is even. Sufficient.
Answer: C
