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# if M and N are integers, is (10^M + N)/3 an integer? 1. N =

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Senior Manager
Joined: 17 Jul 2009
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Concentration: Nonprofit, Strategy
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Kudos [?]: 34 [1] , given: 9

if M and N are integers, is (10^M + N)/3 an integer? 1. N = [#permalink]  06 Aug 2009, 19:56
1
KUDOS
00:00

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(N/A)

Question Stats:

0% (00:00) correct 100% (00:08) wrong based on 3 sessions
if M and N are integers, is (10^M + N)/3 an integer?

1. N = 5
2. MN is even
Manager
Joined: 18 Jul 2009
Posts: 170
Location: India
Schools: South Asian B-schools
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Kudos [?]: 64 [1] , given: 37

Re: is it an integer? [#permalink]  06 Aug 2009, 21:59
1
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....

(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
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Bhushan S.
If you like my post....Consider it for Kudos

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Kudos [?]: 2091 [1] , given: 359

Re: is it an integer? [#permalink]  06 Aug 2009, 22:13
1
KUDOS
Expert's post
Let's consider 10^M

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.

Judging by http://en.wikipedia.org/wiki/Even_and_odd_numbers negative integers can be classified as odd/even. So, our answer is insufficient.

E
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Kudos [?]: 107 [0], given: 39

Re: is it an integer? [#permalink]  06 Aug 2009, 23:14
sprtng wrote:
if M and N are integers, is (10^M + N)/3 an integer?

1. N = 5
2. MN is even

Walker got it , great job
Senior Manager
Joined: 17 Jul 2009
Posts: 299
Concentration: Nonprofit, Strategy
GPA: 3.42
WE: Engineering (Computer Hardware)
Followers: 1

Kudos [?]: 34 [0], given: 9

Re: is it an integer? [#permalink]  07 Aug 2009, 05:33
Intern
Joined: 07 Dec 2009
Posts: 15
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Kudos [?]: 5 [0], given: 4

Re: is it an integer? [#permalink]  17 Dec 2009, 19:25
how did you get 3k + 1? can someone please explain? thanks.
Manager
Joined: 12 Oct 2008
Posts: 58
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Kudos [?]: 1 [0], given: 3

Re: is it an integer? [#permalink]  18 Dec 2009, 20:18
10^K = 3k+1: for any K>=0
It means any number which is 10^K i.e. 1,10,100,1000 ..., when divide by 3, gives remainder 1.
Intern
Joined: 07 Dec 2009
Posts: 15
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Kudos [?]: 5 [0], given: 4

Re: is it an integer? [#permalink]  18 Dec 2009, 20:30
thanks, obviously i still have a lot of catching up to do when it comes to math fundamentals.
Intern
Joined: 17 Jan 2010
Posts: 33
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Kudos [?]: 8 [0], given: 8

Re: is it an integer? [#permalink]  22 Jan 2010, 23:24
yeah, i made the same mistake of opting for A.

but it's E, because we don't no whether m and n are +ve integers
Manager
Joined: 26 Nov 2009
Posts: 178
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Kudos [?]: 53 [0], given: 5

Re: is it an integer? [#permalink]  23 Jan 2010, 09:56
Thanks I committed 2 careless mistakes

Initially I chose A ignoring the fact M and are just integers not positive integers

and then I chose C since MN is even I just assumed it is positive.

but the correct ans is E since both statements would be insufficient to ans the question
Re: is it an integer?   [#permalink] 23 Jan 2010, 09:56
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