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If m and n are both two digit numbers and m-n = 11x, is x an

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If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 30 May 2012, 12:36
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If m and n are both two digit numbers and m-n = 11x, is x an integer?

(1) The tens digit and the units digit of m are same
(2) m+n is a multiple of 11
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 31 May 2012, 01:24
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If m and n are both two digit numbers and m-n = 11x, is x an integer?

The question basically asks whether m-n is a multiple of 11.

(1) The tens digit and the units digit of m are same --> m could be: 11, 22, 33, ..., 99 --> m is a multiple of 11. Not sufficiient since no info about n.

(2) m+n is a multiple of 11 --> if m=n=11 then the m-n is a multiple of 11 but if m=12 and n=10 then m-n is NOT a multiple of 11. Not sufficient.

(1)+(2) From (1) we have that m={multiple of 11} and from (2) we have that m+n={multiple of 11} --> {multiple of 11}+n={multiple of 11} --> n={multiple of 11} --> m-n={multiple of 11}-{multiple of 11}={multiple of 11}. Sufficient.

Answer: C.

Below might help to understand this concept better.

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

Hope it's clear.
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 10 Sep 2013, 21:25
Hi Banuel,

one confusion here.

from (1) we know that m is a multiple of 11. we also know that m-n= multiple of 11.
now, if we consider m to be 99 than , 99-n=multiple of 11. can we have any other 2 digit no. which is NOT a multiple of 11 for n in this case ? I think no. so effectively shouldn't the answer be A ? Have I missed something here ?
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 10 Sep 2013, 23:17
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vishalrastogi wrote:
Hi Banuel,

one confusion here.

from (1) we know that m is a multiple of 11. we also know that m-n= multiple of 11.
now, if we consider m to be 99 than , 99-n=multiple of 11. can we have any other 2 digit no. which is NOT a multiple of 11 for n in this case ? I think no. so effectively shouldn't the answer be A ? Have I missed something here ?


The highlighted portion is the problem. When you say (m-n) is a multiple of 11, that implies that you ARE saying that x IS an integer. But that is something which is in-fact being asked. You can't assume it while solving for the question stem.
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 16 Feb 2014, 08:52
First we know that m is a a multiple of 11 but we still don’t know anything about 'n' therefore insufficient. Then we know that m+n is a multiple of 11 but that doesn't mean that they are both multiples of 11, it could be that they are both not non multiples of 11. Both together since m is a multiple of 11 then n must also be a multiple of 11. The difference of two multiples of 11 is always a multiple of 11. Thus answer is C

Hope this clarifies

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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 11 Apr 2015, 19:10
how can we say that m is a multiple of 11

m could be 11.10
No where mentioned is the stem that M is an int , its just a number

OA should E
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 12 Apr 2015, 03:30
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If m and n are both 2 digit numbers and m-n=11x, is x an integer? [#permalink] New post 20 May 2015, 11:06
If m and n are both 2 digit numbers and m-n=11x, is x an integer?
1)The tens digit and units digit of m are the same
2) m+n is a multiple of 11.
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Re: If m and n are both 2 digit numbers and m-n=11x, is x an integer? [#permalink] New post 20 May 2015, 17:01
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If a number is a multiple of 11, that means you can write the number as '11q' where q is an integer. So if m - n = 11x, and x is an integer, that just means that m-n is a multiple of 11. So that's the question we're trying to answer: is m - n divisible by 11?

Statement 1 tells us that the two-digit number m looks something like 11, 22, 33, etc. Those numbers are all multiples of 11, so Statement 1 is just telling us that m alone is divisible by 11. We know nothing about n, however, so we can't say if m - n is divisible by 11.

Statement 2 tells us that m + n is divisible by 11. This is not sufficient, as you can see by generating almost any two examples, one using multiples of 11, and one not. So maybe m = 22 and n = 11, and then m-n is also divisible by 11. But maybe m = 23 and n = 10, and then m - n is not divisible by 11.

Using both statements, we know that m+n and m are both multiples of 11. That guarantees that n is also a multiple of 11. And if m and n are both multiples of 11, then m-n will always be a multiple of 11 as well, so the answer is C.

If it's unclear why n must be a multiple of 11 here, you can see that in one of the following ways:

• perhaps the fastest is to use the fact that, if we subtract one multiple of 11 from another, we always get a multiple of 11. So if (m+n) and (m) are both multiples of 11, then (m+n) - m will be too, but that's just equal to n.

• the longer way is conceptually more useful to understand, because it illustrates why it's true that you always get a multiple of 11 when you add or subtract two multiples of 11:

- if m+n is a multiple of 11, then m+n = 11q for some integer q
- if m is a multiple of 11, then m = 11k for some integer k

So we know:

m + n = 11q

but m = 11k, so we can substitute for m and rearrange:

11k + n = 11q
n = 11q - 11k
n = 11 (q - k)

and we can now see that n is equal to 11 times some integer, so n is a multiple of 11 also.

Of course there's nothing special about '11' here. When you add or subtract two multiples of any integer p, you always get a multiple of p.
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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink] New post 21 May 2015, 02:33
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Mechmeera wrote:
If m and n are both 2 digit numbers and m-n=11x, is x an integer?
1)The tens digit and units digit of m are the same
2) m+n is a multiple of 11.


Merging similar topics. Please refer to the discussion above.
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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

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Re: If m and n are both two digit numbers and m-n = 11x, is x an   [#permalink] 21 May 2015, 02:33
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