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m17 #8 [#permalink] New post 05 Dec 2008, 18:08
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m and n are positive integers. Is the remainder of \frac{10^m + n}{3} bigger than the remainder of \frac{10^n + m}{3} ?

1. m \gt n
2. The remainder of \frac{n}{3} is 2

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B

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In expression 10^X + Y , the sum of digits of 10^X is always 1 and it is the value of Y that determines the remainder of \frac{10^X + Y}{3} . We can make sure that S1 is not sufficient by plugging m = 2 , n = 1 (the answer is YES) and m = 3 , n = 2 (the answer is NO).

If the remainder of \frac{n}{3} is 2, as S2 states, then n is 2, 5, or 8 and the sum of digits of \frac{10^m + n}{3} is divisible by 3. Therefore, the remainder of \frac{10^m + n}{3} is 0, which cannot be bigger that the remainder of \frac{10^n + m}{3} no matter what m is.
The correct answer is B.

When we evaluate S2, isn't there a possibility that m = n in which case the remainders will be equal? Also, if we take S1+S2, there is a possibility that m/3 has a remainder of 2 as well. IMO the correct answer should be E. Any thoughts?
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Re: m17 #8 [#permalink] New post 06 Dec 2008, 10:50
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you have to read the question really hard, with b the remainder of (10^m + N)/3 has to be zero, meaning m/3's remainder can only be equal or more than, therefore, (10^m + N)/3 's remainder cannot be bigger than (10^n + M)/3, only equal. So the answer to the question is no, regardless of if they are equal or not. Very tricky question, answer definitely helped


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Re: m17 #8 [#permalink] New post 06 Dec 2008, 15:31
sharmar wrote:
Quote:
m and n are positive integers. Is the remainder of \frac{10^m + n}{3} bigger than the remainder of \frac{10^n + m}{3} ?

1. m \gt n
2. The remainder of \frac{n}{3} is 2

(C) 2008 GMAT Club - m17#8

In expression 10^X + Y , the sum of digits of 10^X is always 1 and it is the value of Y that determines the remainder of \frac{10^X + Y}{3} . We can make sure that S1 is not sufficient by plugging m = 2 , n = 1 (the answer is YES) and m = 3 , n = 2 (the answer is NO).

If the remainder of \frac{n}{3} is 2, as S2 states, then n is 2, 5, or 8 and the sum of digits of \frac{10^m + n}{3} is divisible by 3. Therefore, the remainder of \frac{10^m + n}{3} is 0, which cannot be bigger that the remainder of \frac{10^n + m}{3} no matter what m is.
The correct answer is B.


When we evaluate S2, isn't there a possibility that m = n in which case the remainders will be equal? Also, if we take S1+S2, there is a possibility that m/3 has a remainder of 2 as well. IMO the correct answer should be E. Any thoughts?



B. Since (10^m)/3 has always 1 reminder and n/3 has 2 reminder. So \frac{10^m + n}{3} has 0 reminder. Therefore, the reminder of \frac{10^m + n}{3} cannot be greater than the reminder of \frac{10^n + m}{3}.
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Re: m17 #8 [#permalink] New post 28 Jan 2010, 06:18
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\frac{10^m}{3} and \frac{10^n}{3} will always have the same remainder. and if 2 fractions are added then there remainders can also be added as long as excess remainder is adjusted.

So now \frac{m}{3} and \frac{n}{3} will actually decide which remainder is bigger.

S1. M > N doesn't help coz if M = 3 and N 2 then remainder is 1 and if M = 6 and N = 2 or 3 then remainder is 0.
S2. The remainder of N/3 is 2 ===> since the remainder of any integer divided by N can be any value between 0 to N-1 but can never be equal to or greater than N. Hence for any value of M, M/3 will never have remainder greater than 2. Hence Sufficient since remainder of \frac {10^N + M}{3} will consist of X + 0 or 1 or 2 and \frac {10^M + N}{3} consists of X + 2. So 1^st expression can never have remainder greater than the 2^nd expression.
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Re: m17 #8 [#permalink] New post 28 Jan 2010, 10:34
I think ans is C, B doesn't eliminate the scenario when m=n; its not mentioned in question if they are distinct number so we need A m>n so I think ans should be C

what do you think?
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Re: m17 #8 [#permalink] New post 28 Jan 2010, 11:41
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Question: Is Remainder of (10m + n)/3 bigger than Remainder of (10n + m)/3?

Since m and n are positive integers: Remainder of 10m/3 = 1 and Remainder of 10n/3 = 1

Thus: Simplifying the question:

1 + Remainder of (n/3) bigger than 1 + Remainder of (m/3)

=> Remainder of (n/3) bigger than Remainder of (m/3)

(1) m > n Not Sufficient
n = 4; m = 5; R of (n/3) = 1; R of (m/3) = 2 => Yes
n = 4; m = 6; R of (n/3) = 1; R of (m/3) = 0 => No

Remainder of (n/3) = 2 => Not Sufficient
n = 5; m = 5; R of (n/3) = 2; R of (m/3) = 2 => No
n = 8; m = 6; R of (n/3) = 2; R of (m/3) = 0 => Yes

For detailed examples, see the attached excel sheet.

(1): m > n is not sufficient.
(2): Remainder of n/3 = 2 is not sufficient.

I think the answer is E. Your thoughts?
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Re: m17 #8 [#permalink] New post 28 Jan 2010, 12:09
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It's clear, that 1st stmt is insufficient.
The second stmt means, that n = 3k +2, where k is natural.
10^m = 99...9 +1 (the number of 9's is m).
So 10^m + n = 99...9 + 1 + 3k + 2 = 3( 33...3 + k + 1) is divisible by 3 and the ramainder is 0 which cannot bigger than 0, 1 or 2 - the posiible remaiders by division by 3.
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Re: m17 #8 [#permalink] New post 01 Feb 2010, 11:54
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very nice question.

1. m > n doesn't give any sufficient information, to prove this you can assume values.

2. n/3 has remainder 2, that means \frac{(10^m +n)}{3} will always have remainder as 0.
Thus this must be smaller than the other value given, as 0 can be the smallest remainder.
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Re: m17 #8 [#permalink] New post 04 Feb 2010, 08:34
beautiful Question...

even i wasn't getting that why the answer is B instead of E..
but thank you guys for such detailed explanation...
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Re: m17 #8 [#permalink] New post 07 May 2010, 13:47
I can assure one thing, we might not learn any new concept by these questions but will certainly learn a lot from our mistakes.

beauty of these type of questions is, they just asked about the equality/inequality i.e. whether 1st is greater than 2nd or not.

Now consider a case when m/3 also gives remainder 2, most of us will feel since remainder is 0, both are equal thus we cannot solve this question as we have two cases.

even if 2nd gives remainder as 0, the answer to the question remains NO....

Golden Rule : We take so much time in solving the question , always spare 5 seconds to verify what exactly has been asked in DS question- whether it is yes/no question or value question.
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Re: m17 #8 [#permalink] New post 03 Oct 2010, 12:43
(10 ^m + n)/3 , (10^n +m)/3
I'll be using modular arithmetic notation:
10 = 1 mod(3)
100 = 1 (mod3)
1000 = 1 (mod3)
hence 10 ^m,n = 1 (mod3)
problem boils down to n/3 and m/3
remainder can be 0,1,or 2. n=2 (mod3) + the Rem 1 from 1st term implies total rem=3 or 0. B
Lets see if 1 is sufficient.
according to 1:
m = R1 mod(3)
n = R2 mod(3)
m > n doesnt help. Ex: 7, 5 remainders (1,2) ; 8,4 remainders (2,1)
Hence B.
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Re: m17 #8 [#permalink] New post 02 Feb 2011, 09:40
I think it should be C because m can be equal to n. The question just says m and n are positive integers, doesn't say they are unique.
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Re: m17 #8 [#permalink] New post 02 Feb 2011, 10:22
pakba wrote:
I think it should be C because m can be equal to n. The question just says m and n are positive integers, doesn't say they are unique.


You have a valid doubt but still the answer is B.
But I can bet you, you will learn something good today.

check
m17-73674.html#p722035
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Re: m17 #8 [#permalink] New post 05 Feb 2011, 09:25
dmetla wrote:
I think ans is C, B doesn't eliminate the scenario when m=n; its not mentioned in question if they are distinct number so we need A m>n so I think ans should be C

what do you think?


it is B....i also thought it is C but if you read carefully you ll realise it is not asking "which one is bigger?"....

the question is can it be bigger? the answer using B is NO because even if m=n then none is bigger than the other ;-)
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Re: m17 #8 [#permalink] New post 14 Feb 2011, 21:21
GMAT TIGER wrote:
sharmar wrote:
Quote:
m and n are positive integers. Is the remainder of \frac{10^m + n}{3} bigger than the remainder of \frac{10^n + m}{3} ?

1. m \gt n
2. The remainder of \frac{n}{3} is 2

(C) 2008 GMAT Club - m17#8

In expression 10^X + Y , the sum of digits of 10^X is always 1 and it is the value of Y that determines the remainder of \frac{10^X + Y}{3} . We can make sure that S1 is not sufficient by plugging m = 2 , n = 1 (the answer is YES) and m = 3 , n = 2 (the answer is NO).

If the remainder of \frac{n}{3} is 2, as S2 states, then n is 2, 5, or 8 and the sum of digits of \frac{10^m + n}{3} is divisible by 3. Therefore, the remainder of \frac{10^m + n}{3} is 0, which cannot be bigger that the remainder of \frac{10^n + m}{3} no matter what m is.
The correct answer is B.


When we evaluate S2, isn't there a possibility that m = n in which case the remainders will be equal? Also, if we take S1+S2, there is a possibility that m/3 has a remainder of 2 as well. IMO the correct answer should be E. Any thoughts?



B. Since (10^m)/3 has always 1 reminder and n/3 has 2 reminder. So \frac{10^m + n}{3} has 0 reminder. Therefore, the reminder of \frac{10^m + n}{3} cannot be greater than the reminder of \frac{10^n + m}{3}.



I agree that m can equal n. Therefore the answer must be C as prompt one makes equivalency impossible.
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Re: m17 #8 [#permalink] New post 02 Jul 2011, 09:32
I hope GMAC does not hire the person who made this question. :x . Else we are doomed.

Just kidding. Beautiful question again. :)


could not solve this in timed mode, however came up with an elegant solution after having a look at it for 5 min in review.

statement B can be simplified as 3n+2 for any value of N. thus the values at n={1,2,3} are {5,8,11} and so on. Hence the remainder of the 1st expression is always 0. hence the answer.
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Re: m17 #8 [#permalink] New post 16 Jul 2011, 13:14
This is a very nice debatable question. i think answer is B, however my answer in the test was E :(.
In my view question is asking if remainder of 1st expression is greater than remainder of 2nd expression.

So as per 2nd equation, if you choose value of n = 5, 8, 11... and m = 1, 2, 3, 4, 6,... (m <> n)
remainder of 1st expression is less than remainder of 2nd expression.

But if you consider m = n, remainders of both expressions are equal i.e. = 0.
in any case remainder of expression 1 is not greater than remainder of expression 2.

appreciate if someone validate if this understanding is correct?
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Re: m17 #8 [#permalink] New post 09 Feb 2012, 10:03
I understand how statement 2 can be rewritten as 3n+2, leaving R=0. But you know nothing about M, so how can you assume that B is the correct answer?
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Re: m17 #8 [#permalink] New post 09 Feb 2012, 10:33
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pradalove wrote:
I understand how statement 2 can be rewritten as 3n+2, leaving R=0. But you know nothing about M, so how can you assume that B is the correct answer?


m and n are positive integers. Is the remainder of \frac{10^m + n}{3} bigger than the remainder of \frac{10^n + m}{3} ?

First of all any positive integer can yield only three remainders upon division by 3: 0, 1, or 2.

Since, the sum of the digits of 10^m and 10^n is always 1 then the remainders of \frac{10^m + n}{3} and \frac{10^n + m}{3} are only dependant on the value of the number added to 10^m and 10^n. There are 3 cases:
If the number added to them is: 0, 3, 6, 9, ... then the remainder will be 1 (as the sum of the digits of 10^m and 10^n will be 1 more than a multiple of 3);
If the number added to them is: 1, 4, 7, 10, ... then the remainder will be 2 (as the sum of the digits of 10^m and 10^n will be 2 more than a multiple of 3);
If the number added to them is: 2, 5, 8, 11, ... then the remainder will be 0 (as the sum of the digits of 10^m and 10^n will be a multiple of 3).

(1) m \gt n. Not sufficient.

(2) The remainder of \frac{n}{3} is 2 --> n is: 2, 5, 8, 11, ... so we have the third case. Which means that the remainder of \frac{10^m + n}{3} is 0. Now, the question asks whether the remainder of \frac{10^m + n}{3}, which is 0, greater than the reminder of \frac{10^n + m}{3}, which is 0, 1, or 2. Obviously it cannot be greater, it can be less than or equal to. So, the answer to the question is NO. Sufficient.

Answer: B.

Hope it's clear.
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Re: m17 #8 [#permalink] New post 27 Jan 2014, 12:08
Perfect question.
Re: m17 #8   [#permalink] 27 Jan 2014, 12:08
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