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M17-08

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M17-08  [#permalink]

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New post 16 Sep 2014, 00:00
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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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Re M17-08  [#permalink]

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New post 16 Sep 2014, 00:00
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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.


Answer: B
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Re M17-08  [#permalink]

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New post 07 Jan 2015, 08:17
I think this question is good and helpful.
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Re: M17-08  [#permalink]

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New post 11 Feb 2017, 10:26
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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Re: M17-08  [#permalink]

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New post 11 Feb 2017, 11:03
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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Re: M17-08  [#permalink]

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New post 20 Jun 2017, 12:05
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.

please reply, for any error i made.

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Re: M17-08  [#permalink]

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New post 20 Jun 2017, 12:24
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.

please reply, for any error i made.

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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Re: M17-08  [#permalink]

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New post 04 Jul 2017, 11:33
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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Re M17-08  [#permalink]

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New post 19 Apr 2018, 06:03
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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Re: M17-08  [#permalink]

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New post 20 Apr 2018, 01:19
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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Re M17-08  [#permalink]

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New post 17 Jul 2018, 07:55
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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Re M17-08  [#permalink]

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New post 12 Sep 2018, 01:20
I think this is a high-quality question and I agree with explanation. Quite helpful.
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Re M17-08 &nbs [#permalink] 12 Sep 2018, 01:20
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