sidagar wrote:
Hi brunel
evem if M is -2 remainder will be 1 when divided by 3.
10^m = 10^-2 = 1/100 = 1/100/3 = 1/300 gives remainder 1
tell me where i am going wrong .As per me answer mUst by A
The question asks whether (10^M + N)/3 an integer? If M = -2 and N = 5, (10^M + N)/3 = 167/100, which is not an integer.
Below is another solution which might help:
If \(M\) and \(N\) are integers, is \(\frac{10^M + N}{3}\) an integer?Basically the the question ask whether \(10^m+n\) is divisible by 3. Now, in order \(10^m+n\) to be divisible by 3:
A. It must be an integer, and B. the sum of its digits must be multiple of 3.
(1) N = 5 --> if \(m<0\) (-1, -2, ...) then \(10^m+n\) won't be an integer at all (for example if \(m=-1\) --> \(10^m+n=\frac{1}{10}+5=\frac{51}{10}\neq{integer}\)), thus won't be divisible by 3, but if \(m\geq{0}\) (0, 1, 2, ...) then \(10^m+n\) will be an integer and also the sum of its digits will be divisible by 3 (for example for \(m=1\) --> \(10^m+n=10+5=15\) --> 15 is divisible by 3). Not sufficient.
(2) MN is even --> clearly insufficient, as again \(m\) 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
negative even integer (-2, -4, ...), so \(10^m+n\) may or may not be divisible by 3. Not sufficient.
Answer: E.
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