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

Question

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 the same
(2) m + n is a multiple of 11

Bunuel · Accepted Answer

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.

vishalrastogi · Answer

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 ?

mau5 · Answer

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.

jlgdr · Answer

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

Gimme some freaking Kudos!!
Best
J

Hope this clarifies

Gimme some freaking Kudos!!
Best
J

vipulgoel · Answer

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

Bunuel · Answer

11.10 is not a 2-digit number.

IanStewart · Answer

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.

h31337u · Answer

My way to solve this problem was plugging numbers.

(1) m could be 11, 22, 33 ......
If n is 21, then m-n = 1. 
11x = 1 when x is 1/11. So x is not integer. 
If n is 11 and m is 22,
Then m-n = 11
If this is the case, x = 1. So x is integer.
Two different answers. Not sufficient.

(2) 
If m = 22 and n = 11, then m-n = 11. X is integer. 
If m = 23 and n = 10, then m-n = 13. X is not integer (11/13). 
Not sufficient.

(1) + (2)

If m = 22, then n can only be a multiple of 11. Otherwise, (2) can’t be true. 
Also, if m=22, then 
22 + n = 11a
N = 11a-22
N = 11(a-2)
So n should be a multiple of 11.

So, M = 11a n = 11b. (A,b are just randomly chosen variables which will be replaced by positive integers, here it is positive because it is two digit numbers)
11a-11b = 11(a-b)
11(a-b) = 11x
A-b = x
Since, a and b are integers, x must be integers.

IMO = C.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards