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16 Sep 2014, 01:23
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If \(m\) and \(n\) are integers, is \(\frac{10^m + n}{3}\) an integer?
(1) \(n = 5\).
(2) \(mn\) is an even number.
(1) \(n = 5\).
(2) \(mn\) is an even number.
While solving such DS questions always try the ZONEF numbers:
Zero
One
Negatives
Extremes (like 100)
Fractions
These are the numbers that most people forget to think about.
Zero
One
Negatives
Extremes (like 100)
Fractions
These are the numbers that most people forget to think about.
16 Sep 2014, 01:23
Kudos
Bookmarks
Official Solution:
If \(m\) and \(n\) are integers, is \(\frac{10^m + n}{3}\) an integer?
Essentially, the question asks whether \(10^m+n\) is divisible by 3. For \(10^m+n\) to be divisible by 3:
A. It must be an integer, and B. the sum of its digits must be a multiple of 3.
(1) n = 5. If \(m < 0\) (-1, -2, ...), then \(10^m+n\) will not be an integer at all (for example, if \(m=-1\), then \(10^m+n=\frac{1}{10}+5=\frac{51}{10}\neq{integer}\)), thus it won't be divisible by 3. But if \(m \geq{0} \) (0, 1, 2, ...), then \(10^m+n\) will be an integer, and the sum of its digits will also be divisible by 3 (for example, for \(m=1\), then \(10^m+n=10+5=15\), which is divisible by 3). Not sufficient.
(2) \(mn\) is an even number.
Clearly insufficient, as \(m\) can be -2 and \(n\) any integer, and the answer to the question will be NO, or \(m\) can be 0 and \(n\) can be 2, and the answer to the question will be YES. Not sufficient.
(1)+(2) From \(mn=even\) and \(n=5\), it's still possible for \(m\) to be a negative even integer (-2, -4, ...), so \(10^m+n\) may or may not be divisible by 3. Not sufficient.
Answer: E
If \(m\) and \(n\) are integers, is \(\frac{10^m + n}{3}\) an integer?
Essentially, the question asks whether \(10^m+n\) is divisible by 3. For \(10^m+n\) to be divisible by 3:
A. It must be an integer, and B. the sum of its digits must be a multiple of 3.
(1) n = 5. If \(m < 0\) (-1, -2, ...), then \(10^m+n\) will not be an integer at all (for example, if \(m=-1\), then \(10^m+n=\frac{1}{10}+5=\frac{51}{10}\neq{integer}\)), thus it won't be divisible by 3. But if \(m \geq{0} \) (0, 1, 2, ...), then \(10^m+n\) will be an integer, and the sum of its digits will also be divisible by 3 (for example, for \(m=1\), then \(10^m+n=10+5=15\), which is divisible by 3). Not sufficient.
(2) \(mn\) is an even number.
Clearly insufficient, as \(m\) can be -2 and \(n\) any integer, and the answer to the question will be NO, or \(m\) can be 0 and \(n\) can be 2, and the answer to the question will be YES. Not sufficient.
(1)+(2) From \(mn=even\) and \(n=5\), it's still possible for \(m\) to be a negative even integer (-2, -4, ...), so \(10^m+n\) may or may not be divisible by 3. Not sufficient.
Answer: E
