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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

1

This post was BOOKMARKED

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, 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, 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". _________________

Re: If m and z are integers, is z divisible by 11 ? [#permalink]
17 Feb 2015, 01:15

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