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# M17-08

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:00
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Difficulty:

95% (hard)

Question Stats:

35% (01:27) correct 65% (01:50) wrong based on 129 sessions

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If $$m$$ and $$n$$ are positive integers, is the remainder of $$\frac{10^m + n}{3}$$ greater 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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16 Sep 2014, 01:00
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1
Official Solution:

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 dependent 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 $$2$$. So, $$n$$ could be: 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, is 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.

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07 Jan 2015, 09:17
I think this question is good and helpful.
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Hasan Mahmud
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11 Feb 2017, 11:26
Hi Bunuel

For question stem ,why can't we do below steps :

$$\frac{10^m}{3}$$ + $$\frac{n}{3}$$ > $$\frac{10^n}{3}$$+ $$\frac{m}{3}$$

cancelling $$\frac{10^m}{3}$$ and $$\frac{10^n}{3}$$

==> $$\frac{n}{3}$$ > $$\frac{m}{3}$$

==> n > m

Stat 1 says m > n. Sufficient .

Thanks
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11 Feb 2017, 12:03
pranjal123 wrote:
Hi Bunuel

For question stem ,why can't we do below steps :

$$\frac{10^m}{3}$$ + $$\frac{n}{3}$$ > $$\frac{10^n}{3}$$+ $$\frac{m}{3}$$

cancelling $$\frac{10^m}{3}$$ and $$\frac{10^n}{3}$$

==> $$\frac{n}{3}$$ > $$\frac{m}{3}$$

==> n > m

Stat 1 says m > n. Sufficient .

Thanks

$$\frac{10^[highlight]m[}{highlight]/3}$$
$$\frac{10^[highlight]n[}{highlight]/3}$$

Those two expressions are not the same, so you cannot cancel them.
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20 Jun 2017, 13:05
Hi,

i think statement 1 is also sufficient. if we do binomial expansion of 10^m as (9+1)^m. only last term will be 1^m and because m is positive integer remainder will always be 1 when 10^m is divided by 3. we will be left with (1+n)/3 in LHS and (1+m)/3 on RHS. hence we only need to prove weather m>n or not.

statement 1 also seems sufficient.

Regards

Sushil kumar
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20 Jun 2017, 13:24
Sushilait84 wrote:
Hi,

i think statement 1 is also sufficient. if we do binomial expansion of 10^m as (9+1)^m. only last term will be 1^m and because m is positive integer remainder will always be 1 when 10^m is divided by 3. we will be left with (1+n)/3 in LHS and (1+m)/3 on RHS. hence we only need to prove weather m>n or not.

statement 1 also seems sufficient.

Regards

Sushil kumar

It seems that you did not read the solution above. Please re-read it carefully. If still not satisfied please visit this discussion: https://gmatclub.com/forum/if-m-and-n-a ... 01636.html. Apart of several different solutions, you can find the one with binomial approach: https://gmatclub.com/forum/if-m-and-n-a ... l#p1073179

Hope it helps.
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04 Jul 2017, 12:33
1
One could alternatively solve this question by testing some values

So WKT m,n are positive integers. Question - Is rem of $$\frac{(10^m + n)}{3}$$ > rem of $$\frac{(10^n + m)}{3}$$? Yes/No

1) m>n
Say m = 2 and n = 1
rem of $$\frac{(10^2 + 1)}{3}$$ = 2
rem of $$\frac{(10^1 + 2)}{3}$$ = 0
YES

[Trying to disprove]
Say m = 3 and n = 2
rem of $$\frac{(10^3 + 2)}{3}$$ = 0
rem of $$\frac{(10^2 + 3)}{3}$$ = 1
NO

Hence Insuff

2) This is interesting n/3 = __r2
n = 2, 5, 8, 11....
Notice how this effects the rem of $$\frac{(10^m + n)}{3}$$. No matter what you plug in for m the rem is ALWAYS 0. And since the rem CANNOT be neg numbers, rem of $$\frac{(10^n + m)}{3}$$ can be 0 or greater, leading to ALWAYS NO answer.
Hence suff.
Option B
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19 Apr 2018, 07:03
I think this is a high-quality question and I agree with explanation. I am not able to correctly view some of the calculations posted in the comment section.
Do i need to change something as it is showing digits superimposed on eavh other.
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20 Apr 2018, 02:19
luffy92 wrote:
I think this is a high-quality question and I agree with explanation. I am not able to correctly view some of the calculations posted in the comment section.
Do i need to change something as it is showing digits superimposed on eavh other.

Try to reload the page couple of times. If it does not help, please post a screenshot and we'll try to find out.
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17 Jul 2018, 08:55
I think this is a high-quality question and I agree with explanation. Given, m,n are positive numbers. So taking '0' in to calculation is wrong, I guess. In the step... value of m,n be 0,3,6,9,.................
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12 Sep 2018, 02:20
I think this is a high-quality question and I agree with explanation. Quite helpful.
Re M17-08   [#permalink] 12 Sep 2018, 02:20
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# M17-08

Moderators: chetan2u, Bunuel