Author 
Message 
Current Student
Joined: 26 Jun 2006
Posts: 124
Schools: Ross Class of 2012
WE 1: Consumer Goods / IT
WE 2: Retail / Logistics

7
This post received KUDOS
5
This post was BOOKMARKED
\(m\) 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 \(2\) Source: GMAT Club Tests  hardest GMAT questions 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?



Manager
Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 209
Schools: Harvard, Penn, Maryland

Re: m17 #8 [#permalink]
Show Tags
06 Dec 2008, 11:50
2
This post received KUDOS
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
B



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: m17 #8 [#permalink]
Show Tags
06 Dec 2008, 16:31
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 \(\frac{10^n + m}{3}\).
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Intern
Joined: 18 Jan 2010
Posts: 17

Re: m17 #8 [#permalink]
Show Tags
28 Jan 2010, 07:18
1
This post received KUDOS
\(\frac{10^m}{3} and \frac{10^n}{3}\) 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 N1 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.



Manager
Joined: 27 Aug 2009
Posts: 137

Re: m17 #8 [#permalink]
Show Tags
28 Jan 2010, 11:34
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?



Intern
Joined: 11 Jan 2010
Posts: 38

Re: m17 #8 [#permalink]
Show Tags
28 Jan 2010, 12:41
1
This post received KUDOS
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?



Intern
Joined: 15 Nov 2009
Posts: 31
Location: Moscow, Russia

Re: m17 #8 [#permalink]
Show Tags
28 Jan 2010, 13:09
1
This post received KUDOS
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. B



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: m17 #8 [#permalink]
Show Tags
01 Feb 2010, 12:54
1
This post received KUDOS
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.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 19 Feb 2009
Posts: 55
Schools: INSEAD,Nanyang Business school, CBS,

Re: m17 #8 [#permalink]
Show Tags
04 Feb 2010, 09:34
beautiful Question... even i wasn't getting that why the answer is B instead of E.. but thank you guys for such detailed explanation...
_________________
Working without expecting fruit helps in mastering the art of doing faultfree action !



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: m17 #8 [#permalink]
Show Tags
07 May 2010, 14:47
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.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Senior Manager
Joined: 29 Sep 2009
Posts: 392

Re: m17 #8 [#permalink]
Show Tags
03 Oct 2010, 13:43
(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.



Intern
Joined: 01 Feb 2011
Posts: 1

Re: m17 #8 [#permalink]
Show Tags
02 Feb 2011, 10:40
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.



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: m17 #8 [#permalink]
Show Tags
02 Feb 2011, 11:22
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 m1773674.html#p722035
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 16 Feb 2010
Posts: 225

Re: m17 #8 [#permalink]
Show Tags
05 Feb 2011, 10:25
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



Manager
Joined: 16 Sep 2010
Posts: 220
Location: United States
Concentration: Finance, Real Estate

Re: m17 #8 [#permalink]
Show Tags
14 Feb 2011, 22:21
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.



Manager
Affiliations: The Earth organization, India
Joined: 25 Dec 2010
Posts: 191
WE 1: SAP consultantIT 2 years
WE 2: Entrepreneurfamily business 2 years

Re: m17 #8 [#permalink]
Show Tags
02 Jul 2011, 10:32
I hope GMAC does not hire the person who made this question. . 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.
_________________
Cheers !!
Quant 47Striving for 50 Verbal 34Striving for 40



Senior Manager
Joined: 24 Mar 2011
Posts: 447
Location: Texas

Re: m17 #8 [#permalink]
Show Tags
16 Jul 2011, 14:14
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?



Intern
Joined: 31 Jan 2012
Posts: 2

Re: m17 #8 [#permalink]
Show Tags
09 Feb 2012, 11:03
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?



Math Expert
Joined: 02 Sep 2009
Posts: 39678

Re: m17 #8 [#permalink]
Show Tags
09 Feb 2012, 11:33
5
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 03 Jun 2012
Posts: 83
Location: United States
GPA: 2.7
WE: Analyst (Investment Banking)

Re: m17 #8 [#permalink]
Show Tags
27 Jan 2014, 13:08
Perfect question.











