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If m and n are positive integers, is the remainder of

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If \(m\) and \(n\) are positive integers, is the remainder of \(\frac{10^m + n}{3}\) larger than the remainder of \(\frac{10^n + m}{3}\) ?

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

[Reveal] Spoiler:
i disagree with the OA answer: -

"Statement (2) by itself is sufficient. If the remainder of \(\frac{n}{3}\) is 2, as S2 states, then \(n\) is 2, 5, or 8 and the sum of the 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 larger that the remainder of \(\frac{10^n + m}{3}\) no matter what \(m\) is."but if m=n, then the remainder of both expresions will be 0!
even more weird is if m=2, n=5, therefore remainder 0 for both....


please advice
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New post 24 Sep 2010, 14:42
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The question asks if the reminder can be larger though. So even if the m = n and the reminders of both terms are 0, it's still not larger.

I agree with answer B.
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To determine if a number is divisible by 3, add the digits and if they are divisible by 3, then the entire number is also.

We are given m and n are positive integers. The sum of the digits of 10 to any positive power will always be 1.

From statement 2, we are told the remainder of \(n/3\) will be 2.

For \((10^m + n)/3\), the \(10^m\) part will always give you 1 in the sum of the digits, and the n part will always give you 2 in the digits of the numerator. Therefore, this will always be divisible by 3 (have a remainder of 0).

The question asked if (10^m + n)/3 will ever have a larger remainder than \((10^n + m)/3\). Remainders can't be negative, therefore we know it will never be larger. There is no need to evaluate possible remainders of \((10^n + m)/3\).
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New post 10 Apr 2012, 07:02
For (10^m + n) /3, the 10^m part will always give you 1 in the sum of the digits, and the n part will always give you 2 in the digits of the numerator. Therefore, this will always be divisible by 3 (have a remainder of 0).

I'm not sure i understand the bold face part, why will it give you 2? how did you come to this conclusion?
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catty2004 wrote:
For (10^m + n) /3, the 10^m part will always give you 1 in the sum of the digits, and the n part will always give you 2 in the digits of the numerator. Therefore, this will always be divisible by 3 (have a remainder of 0).

I'm not sure i understand the bold face part, why will it give you 2? how did you come to this conclusion?


\(m\) 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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New post 10 Apr 2012, 07:28
crystal clear! Just hope that my brain will remember this on the test day!
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New post 12 Apr 2012, 10:28
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zisis wrote:
If \(m\) and \(n\) are positive integers, is the remainder of \(\frac{10^m + n}{3}\) larger than the remainder of \(\frac{10^n + m}{3}\) ?

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



You can also use binomial theorem here. Again, let me reiterate that there are many concepts which are not essential for GMAT but knowing them helps you get to the answer quickly.

The moment I see \(\frac{10^m + n}{3}\) here, my mind sees \(\frac{(9+1)^m + n}{3}\)
So I say that \(\frac{10^m}{3}\) and \(\frac{10^n}{3}\) give remainder 1 in any case (m and n are positive integers). I just need to worry about n/3 and m/3.

1. \(m \gt n\)
Doesn't tell me about the remainder when m and n are divided by 3.

2. The remainder of \(\frac{n}{3}\) is \(2\)
If n/2 gives a remainder of 2, total remainder of \(\frac{10^m + n}{3}\) is 1+2 = 3 which is equal to 0. So no matter what the remainder of \(\frac{m}{3}\), the remainder of \(\frac{10^n + m}{3}\) will never be less than 0. Hence sufficient.

Answer B
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Re: If m and n are positive integers, is the remainder of [#permalink]

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New post 29 Apr 2012, 09:32
I have a doubt why B is sufficient to answer the question.
I understood the solution but if we take only the statement B then there is nothing to prove that m is not equal to n.
As in the given statement m and n are positive integers but what is the relation between them is not provided whether m >n , m<n or m=n .

Also ,by B we are only able to prove that remainder will not be lesser but what if equal or greater.
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Re: remainder [#permalink]

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New post 30 Apr 2012, 05:04
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I have a doubt why B is sufficient to answer the question.
I understood the solution but if we take only the statement B then there is nothing to prove that m is not equal to n.
As in the given statement m and n are positive integers but what is the relation between them is not provided whether m >n , m<n or m=n .

Also ,by B we are only able to prove that remainder will not be lesser but what if equal or greater.


The question is:
is the remainder of 'a' larger than the remainder of 'b'?

We found that the remainder of 'a' is 0 which is the smallest possible remainder. No matter what the remainder of 'b' is, it will never be less than 0. So remainder of 'a' can never be larger than the remainder of 'b'. We can answer the question with 'No'. Therefore, statement 2 is sufficient.
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Re: If m and n are positive integers, is the remainder of [#permalink]

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New post 23 Jul 2012, 20:18
to rephrase the question is x greater than y ( here x and y are the remainders).
If x is greater than y, answer is yes but if x is not greater than y the answer is no.

If x is 0 and y is 0 ----No
If x is 1 or 2 and y is 0 , the Yes

Why is the answer not E?

If the question is reversed -- is the remainder of 10^n +m/3 larger than the remainder of 10^m+n/3 ? then the answer would be B as
0 can never be greater than 0, 1 or 2.
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New post 24 Jul 2012, 04:17
nomis wrote:
to rephrase the question is x greater than y ( here x and y are the remainders).
If x is greater than y, answer is yes but if x is not greater than y the answer is no.

If x is 0 and y is 0 ----No
If x is 1 or 2 and y is 0 , the Yes

Why is the answer not E?

If the question is reversed -- is the remainder of 10^n +m/3 larger than the remainder of 10^m+n/3 ? then the answer would be B as
0 can never be greater than 0, 1 or 2.


Because you find that a = 0 (the remainder of the first expression is 0). Hence under no condition can 'a' be greater than 'b'. Hence, it is sufficient to answer with an emphatic 'No'.
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New post 20 Oct 2013, 23:42
karishma I did not understand this statement the total remainder 1+2=3 which is equal to zero
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New post 21 Oct 2013, 09:19
The remainder of 10/3 is 1. So 1^n =1 = 1^m is the remainder of 10^n / 3 as well as 10^m/3. Therefore, the remainder of (10^n + m)/3 is the remainder of (1+m)/3 and the remainder of (10^m +n)/3 is equal to the remainder of (1 +n)/3.
(1) m >n which does not give us any extra information about the remainder of m,n/3.
(2) 2 is the remainder of n/3, so 0 is the remainder of (1+n)/3. That means the remainder of \frac{10^m + n}{3} is not larger than the remainder of \frac{10^n + m}{3}
The answer is B
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Re: If m and n are positive integers, is the remainder of [#permalink]

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New post 14 Nov 2014, 01:43
mohnish104 wrote:
karishma I did not understand this statement the total remainder 1+2=3 which is equal to zero


Even if my reply is belated. You can perform addition/subtraction/multiplication/division with reminders as long as you correct the excess, because a reminder must be at least one unit smaller than a divisor. If an expression yields R1 and the other expression yields R2 the overall expression in this case a sum yields R3 but since R3 is an illegal value you have to correct the excess and you do that by subtracting 3. Now R3-R3=R0 --> multiple of 3. That's what she said *wink.

Regarding the whole question, I'd suggest to write down a few numbers and see the behavior when divided by 3. It is helpful especially with statement 2 because you will be visually able to realize that is talking about (3n-1). Not super hard, but time consuming for sure, at least for me.
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Re: If m and n are positive integers, is the remainder of [#permalink]

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New post 06 Feb 2016, 07:16
Bunuel wrote:
catty2004 wrote:
For (10^m + n) /3, the 10^m part will always give you 1 in the sum of the digits, and the n part will always give you 2 in the digits of the numerator. Therefore, this will always be divisible by 3 (have a remainder of 0).

I'm not sure i understand the bold face part, why will it give you 2? how did you come to this conclusion?


\(m\) 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.



what do the straight line after any number means???
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New post 06 Feb 2016, 10:23
nishantdoshi wrote:
Bunuel wrote:
catty2004 wrote:
For (10^m + n) /3, the 10^m part will always give you 1 in the sum of the digits, and the n part will always give you 2 in the digits of the numerator. Therefore, this will always be divisible by 3 (have a remainder of 0).

I'm not sure i understand the bold face part, why will it give you 2? how did you come to this conclusion?


\(m\) 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.



what do the straight line after any number means???


That's just formatting issue which will be fixed soon.
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Re: If m and n are positive integers, is the remainder of [#permalink]

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New post 06 Feb 2016, 18:23
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If m and n are positive integers, is the remainder of 10m+n/3 larger than the remainder of 10n+m/3 ?

(1) m>n
(2) The remainder of n/3 is 2


Modify the original condition and the question. You need to figure out a remainder dividing n+1 with 3, regardless of the remainder dividing 10^m+n with 3 apart from m. Also, you need to figure out a remainder dividing m+1 with 3, regardless of the remainder dividing 10^n+m with 3 apart from n.
Thus, there are 2 variables, which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), if n=2, m=3, 2+1 divided by 3 is 0 and 3+1 divided by 3 is 1, which is no and sufficient. That makes C the answer.
Since this is an integer question, one of the key questions, apply the mistake type 4(A). In 1) m>n, m=2, n=1 -> yes, m=3, n=2 -> no, which is not sufficient. For 2), if you substitute n=2,5,8.., the remainder is 0 when dividing n+1 with 3. Then, m+1 divided by 3 can have a remainder amongst 0,1,2, which is always no and sufficient. Therefore, the answer is B.
Getting this type of question right can lead you to score 50-51.


-> Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If m and n are positive integers, is the remainder of   [#permalink] 06 Feb 2016, 18:23

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8 Experts publish their posts in the topic If m and n are positive integers such that m > n, what is the remainde Bunuel 7 06 Apr 2015, 06:55
Experts publish their posts in the topic If m and n are positive integers is m/n an integer? sunita123 3 13 Dec 2014, 12:55
5 Experts publish their posts in the topic If n is a positive integer and r is the remainder when (n - ksharma12 6 26 Apr 2010, 13:39
3 n is a positive integer. What is the remainder when n is andrehaui 8 09 Apr 2007, 10:59
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If m and n are positive integers, is the remainder of

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