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# M25-07

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Math Expert
Joined: 02 Sep 2009
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M25-07  [#permalink]

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16 Sep 2014, 01:23
6
20
00:00

Difficulty:

95% (hard)

Question Stats:

36% (00:51) correct 64% (00:51) wrong based on 677 sessions

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If $$M$$ and $$N$$ are integers, is $$\frac{10^M + N}{3}$$ an integer?

(1) $$N = 5$$

(2) $$MN$$ is even

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Re M25-07  [#permalink]

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16 Sep 2014, 01:23
3
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Official Solution:

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 $$M = -2$$; $$N = 5$$.

Answer: E
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Re: M25-07  [#permalink]

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07 Oct 2014, 05:17
1
Amazing. My mind doenst prompt me to thing on the negative number
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Re: M25-07  [#permalink]

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07 Oct 2014, 09:53
@Dennis14:- Even I did not think of the negative numbers.

But I tried substituting values M satisfying MN-even. We have more than one value and so both options are not sufficient.
It took me more than a minute and Bunuel was spot on. 15 seconds max.
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M25-07  [#permalink]

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Updated on: 25 Dec 2014, 14:05
11
15
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.

Originally posted by vfqrk23 on 07 Oct 2014, 11:49.
Last edited by vfqrk23 on 25 Dec 2014, 14:05, edited 1 time in total.
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M25-07  [#permalink]

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20 Dec 2014, 11:03
It was tricky Initially I went for option A then realized M and N are integers not positive integers .
ZONEF !! It seems to be helpful
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Re: M25-07  [#permalink]

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28 Jun 2015, 06:24
E for me.. Yess we can take m=-2, 10^-2 become 1/100 and which will always contradict with positive values.
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Re: M25-07  [#permalink]

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03 Jul 2015, 08:11
Amazing analysis....
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Re: M25-07  [#permalink]

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18 Jul 2016, 01:39
Bunuel wrote:
Official Solution:

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 $$M = -2$$; $$N = 5$$.

Answer: E

no matter how much I train my mind to think of negative scenarios, I forget every time..
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Re: M25-07  [#permalink]

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05 Jul 2017, 01:41
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
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Re: M25-07  [#permalink]

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05 Jul 2017, 02:00
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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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Re: M25-07  [#permalink]

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06 Jun 2018, 10:34
How about ZOSNEF? Add 'Square root' there as well !
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Re M25-07  [#permalink]

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10 Aug 2018, 10:40
I think this is a high-quality question and I don't agree with the explanation.
Re M25-07   [#permalink] 10 Aug 2018, 10:40
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# M25-07

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