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

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If m and z are integers, is z divisible by 11 ? [#permalink]

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06 Feb 2012, 08:11
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If m and z are integers, is z divisible by 11 ?

(1) m is a 2 digit number with equal units and tens digits
(2) m - z is divisible by 11
[Reveal] Spoiler: OA

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Re: Divisibility by 11 [#permalink]

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

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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.

Ans=E

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Re: Divisibility by 11 [#permalink]

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06 Feb 2012, 08:37
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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.

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: Divisibility by 11 [#permalink]

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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

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Re: Divisibility by 11 [#permalink]

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06 Feb 2012, 13:05
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".
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Re: If m and z are integers, is z divisible by 11 ? [#permalink]

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

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14 Sep 2016, 05:58
Hello from the GMAT Club BumpBot!

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Re: If m and z are integers, is z divisible by 11 ?   [#permalink] 14 Sep 2016, 05:58
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# If m and z are integers, is z divisible by 11 ?

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