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Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink]
31 May 2012, 01:24

2

This post received KUDOS

Expert's post

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.

Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink]
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 ?

Re: If m and n are both two digit numbers and m-n = 11x, is x an [#permalink]
10 Sep 2013, 23:17

Expert's post

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. _________________

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

Gimme some freaking Kudos!! Best J

gmatclubot

Re: If m and n are both two digit numbers and m-n = 11x, is x an
[#permalink]
16 Feb 2014, 08:52