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"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....

Re: GMAT Club > Tests > Number Properties - I Question 8 [#permalink]
24 Sep 2010, 14:52

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This post received KUDOS

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.

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

Re: If m and n are positive integers, is the remainder of [#permalink]
10 Apr 2012, 07:08

4

This post received KUDOS

Expert's post

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.

Re: If m and n are positive integers, is the remainder of [#permalink]
12 Apr 2012, 10:28

4

This post received KUDOS

Expert's post

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.

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

Re: If m and n are positive integers, is the remainder of [#permalink]
30 Apr 2012, 05:04

Expert's post

raingary wrote:

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. _________________

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

Re: If m and n are positive integers, is the remainder of [#permalink]
24 Jul 2012, 04:17

Expert's post

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'. _________________

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

gmatclubot

Re: If m and n are positive integers, is the remainder of
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21 Oct 2013, 09:19

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...