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Re: Divisibility by 11 [#permalink]
06 Feb 2012, 08:37

1

This post received KUDOS

Expert's post

kys123 wrote:

Statement 1: Tells nothing about Z. Not enough info Statement 2: Lots of number combination will work. Not enough info

1+2= m= [99,88,77,66,55,44,33,22,11,00,-11....] Z could be the same numbers. If Z = 11,22,33 then it is divisible, but if Z was 0 then it's not divisible. Not enough info.

Ans=E

Zero is divisible by every integer except zero itself: 0/integer=0. _________________

Re: Divisibility by 11 [#permalink]
06 Feb 2012, 08:24

Statement 1: Tells nothing about Z. Not enough info Statement 2: Lots of number combination will work. Not enough info

1+2= m= [99,88,77,66,55,44,33,22,11,00,-11....] Z could be the same numbers. If Z = 11,22,33 then it is divisible, but if Z was 0 then it's not divisible. Not enough info.

Re: Divisibility by 11 [#permalink]
06 Feb 2012, 08:37

Expert's post

If m and z are integers, is z divisible by 11?

(1) m is a 2 digit number with equal units and tens digits --> m is of a type 11, 22, ..., 99 so basically we are told that m is divisible by 11. Though still insufficient, as no info about z.

(2) m - z is divisible by 11. Clearly insufficient.

(1)+(2) {multiple of 11}-z={multiple of 11} --> z={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: Divisibility by 11 [#permalink]
06 Feb 2012, 11:27

Thanks Bunuel. I had no idea that 0 has that type of property where is divisible by any number. I am assuming any number with no remainder after being divided is divisible. Also another question, so let's say the question was, is Z a multiple of 11 instead of divisible by 11 then the answer will be E

Re: Divisibility by 11 [#permalink]
06 Feb 2012, 13:05

Expert's post

kys123 wrote:

Thanks Bunuel. I had no idea that 0 has that type of property where is divisible by any number. I am assuming any number with no remainder after being divided is divisible. Also another question, so let's say the question was, is Z a multiple of 11 instead of divisible by 11 then the answer will be E

a is a multiple of b means that a=kb, for some integer k; a is divisible by b means that a/b=integer;

Which means that a is a multiple of b is the same as a is divisible by b (just remember that 0/0 is undefined).

So, there is no difference between saying "z is divisible by 11" and "z is a multiple of 11". _________________