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If M and N are integers, is (10^M + N)/3 an integer?
(1) N = 5
(2) MN is even
(1) N = 5
(2) MN is even
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24 Nov 2010, 02:44
Kudos
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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
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
General Discussion
06 Aug 2009, 22:59
Kudos
Bookmarks
A
Statement 1: if u take N=5 and any value for M it will always hold true that is (10^M + N)/3 an integer; becoz 10^M ( let M = any value) will always leave a reminder of 1....and 1+5 = 6 which is divisable by 3 ....sufficient
Statement 2 : if MN is even ...this means (M,N) can be both even or odd (except both being odd at a time)....if taken an example....
(M,N) => (1,2) => 10+2 =12/3 => integer
(M,N) => (1,4) => 10+4 =14/3 => non-integer
hence insufficient
if u like my post....consider it for Kudos
Statement 1: if u take N=5 and any value for M it will always hold true that is (10^M + N)/3 an integer; becoz 10^M ( let M = any value) will always leave a reminder of 1....and 1+5 = 6 which is divisable by 3 ....sufficient
Statement 2 : if MN is even ...this means (M,N) can be both even or odd (except both being odd at a time)....if taken an example....
(M,N) => (1,2) => 10+2 =12/3 => integer
(M,N) => (1,4) => 10+4 =14/3 => non-integer
hence insufficient
if u like my post....consider it for Kudos
