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Re: If M and N are integers, is (10^M + N)/3 an integer? [#permalink]

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06 Aug 2009, 22:59

1

This post received KUDOS

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....

at M>=0: \(10^M = {1, 10, 100, 1000....}\) or \(3k+1\) at M<0: 10^M is a fraction and our expression is not an integer.

Now, let's see our statements:

1) N=5 at M>=0: (3k+1 + 5) /3 = k+2 - an integer at M<0: a fraction. insufficient

2) MN is even a) at M=2, N=5 our expression is an integer b) at M=-2, N=-5 our expression in a fraction insufficient

1)&2) N=5 & MN is even --> N=5, M is even For all even M>=0 we will get an integer. Now we have interesting question: Could negative numbers be even or odd? if yes, we can choose M=-2, N=5 and get a fraction. if no, M cannot be negative and two statements are sufficient.

If M and N are integers, is ((10^M)) + N) / 3 an integer?

1. N = 5 2. MN is even

Answer to question is E.

From statement 1 and 2, M has to be Even Positive as N = 5 and MN is even.

Now when 5 is added to 10^M where M is even positive and then divided by 3 it always results into 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.

Re: If M and N are integers, is ((10^M)) + N)/3 an integer? [#permalink]

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19 Sep 2013, 22:22

Hi,

"I read in a forum that negative numbers cannot be tagged as odd or even. So, if I get a DS question stating that X is a even integer, then it also implies that X>=0."

However, in the question given below, the logic is not holding true. I was considering the right answer to be C), because if MN = even and N =5 , then it implies that M must be odd integer (and hence M is > 0). Can you clarify if the argument above is false?

If M and N are integers, is 10M+N3 an integer? (1) N=5

(2) MN is even

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.EACH statement ALONE is sufficient.Statements (1) and (2) TOGETHER are NOT sufficient.Mark as a guessHide Answer The question does not mention whether M and N are positive. Hence, the statements taken together are not sufficient because the answer is YES if M=2, N=5 and NO if M=−2; N=5.

Re: If M and N are integers, is ((10^M)) + N)/3 an integer? [#permalink]

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19 Sep 2013, 23:44

SurabhiStar wrote:

Hi,

"I read in a forum that negative numbers cannot be tagged as odd or even. So, if I get a DS question stating that X is a even integer, then it also implies that X>=0."

However, in the question given below, the logic is not holding true. I was considering the right answer to be C), because if MN = even and N =5 , then it implies that M must be odd integer (and hence M is > 0). Can you clarify if the argument above is false?

If M and N are integers, is 10M+N3 an integer? (1) N=5 (2) MN is even

First, every number except zero "0" can be categorized either Even or odd. Secondly, "X is a even integer" does not necessarily mean X>=0. X can be smaller than 0 as well.
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"I read in a forum that negative numbers cannot be tagged as odd or even. So, if I get a DS question stating that X is a even integer, then it also implies that X>=0."

However, in the question given below, the logic is not holding true. I was considering the right answer to be C), because if MN = even and N =5 , then it implies that M must be odd integer (and hence M is > 0). Can you clarify if the argument above is false?

If M and N are integers, is 10M+N3 an integer? (1) N=5 (2) MN is even

First, every number except zero "0" can be categorized either Even or odd. Secondly, "X is a even integer" does not necessarily mean X>=0. X can be smaller than 0 as well.

1. EVEN/ODD

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder.

An odd number is an integer that is not evenly divisible by 2.

According to the above both negative and positive integers can be even or odd.

2. ZERO

Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

Re: If M and N are integers, is ((10^M)) + N)/3 an integer? [#permalink]

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19 Jan 2015, 00:42

Nice question....

Initially didnt observe that M can be negative....then it looked too easy....well had it come in the early part of real test....i wud have thought im still in easy qtns zone....and put it as A

From statement 1 and 2, M has to be Even Positive as N = 5 and MN is even.

Now when 5 is added to 10^M where M is even positive and then divided by 3 it always results into an integer.

M25-07

Question: Is ((10^M)) + N)/3 an integer? Re-worded: Is (10^M + N) a multiple of 3?

Note that 10^M will be something like 1000... if M is non-negative. Then, 10^M will be of the form (3a+1) because it will leave remainder 1 when divided by 3. If M is negative, 10^M will not be an integer.

Assuming M is non-negative, for 10^M + N to be a multiple of 3, N should be of them form 3n+2.

(1) N = 5 We don't know whether M is non-negative. Not sufficient.

(2) MN is even again, we don't know whether M is non-negative. Not sufficient.

Using both, we know that M is even since N = 5 (odd) but we don't know whether it is non-negative.

From statement 1 and 2, M has to be Even Positive as N = 5 and MN is even.

Now when 5 is added to 10^M where M is even positive and then divided by 3 it always results into an integer.

M25-07

Question: Is ((10^M)) + N)/3 an integer? Re-worded: Is (10^M + N) a multiple of 3?

Note that 10^M will be something like 1000... if M is non-negative. Then, 10^M will be of the form (3a+1) because it will leave remainder 1 when divided by 3. If M is negative, 10^M will not be an integer.

Assuming M is non-negative, for 10^M + N to be a multiple of 3, N should be of them form 3n+2.

(1) N = 5 We don't know whether M is non-negative. Not sufficient.

(2) MN is even again, we don't know whether M is non-negative. Not sufficient.

Using both, we know that M is even since N = 5 (odd) but we don't know whether it is non-negative.

Answer (E)

should the answer not be A as statement one tells us that the expression will be an integer. the question asks if the expression will be an integer and it is clearly visible that the statement 1 with N=5 will not leave a remainder and hence be divisible and hence will be an integer. Is this not sufficient to answer. the expression can also be written as (10^m)/3 + (N/3) = (1^m)/3 + 5/3= remainder of 0 and hence will be an integer. theoretically this should suffice, however, plugging values is a different result. please clear the doubt.

The prompt tells us that M and N are integers. It does NOT state that they are necessarily positive integers though. If you assume that M and N are positive, then your deduction is correct. However, what if M = -1.....? That IS an integer, but it will NOT lead to an integer result when you answer the question.