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M20-08

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16 Sep 2014, 01:08
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If $$x$$ and $$y$$ are positive integers, is the integer $$10^x - y$$ divisible by 9 ?

(1) $$y$$ is divisible by 3

(2) $$10^x + y$$ is not divisible by 9
[Reveal] Spoiler: OA

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16 Sep 2014, 01:08
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Official Solution:

Statement (1) by itself is sufficient. From S1: $$10^x - y$$ = (something not divisible by 3) - (something divisible by 3) = (something not divisible by 3). Because $$10^x - y$$ is not divisible by 3, it is also not divisible by 9.

Statement (2) by itself is insufficient. If $$y = 1$$, the answer is "yes"; if $$y = 2$$, the answer is "no".

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28 Mar 2016, 21:46
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How is statement 1 sufficient?

If we pick numbers for statement 1 we can get two different answers.
if x=2 and y=3, then 100-3=97, Answer NO
if x=2 and y=9, then 100-9=81, Answer YES

Am I missing something?

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28 Mar 2016, 22:16
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oirfan wrote:
How is statement 1 sufficient?

If we pick numbers for statement 1 we can get two different answers.
if x=2 and y=3, then 100-3=97, Answer NO
if x=2 and y=9, then 100-9=81, Answer YES

Am I missing something?

100 - 9 = 91, not 81.
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23 May 2017, 06:21
Dear Sir,

I am having a doubt which might sound silly. Hence my apologies.

From the question, i understand is ((10^x)-y)/9?

Is it okay if I re-write this way: (10^x/9)-y/9 since they will have comment denominator?

Is that case, since x and y are positive, wouldn't the stem of the question itself denotes that it is not divisible by 9?

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23 May 2017, 06:34
Manoraaju wrote:
Dear Sir,

I am having a doubt which might sound silly. Hence my apologies.

From the question, i understand is ((10^x)-y)/9?

Is it okay if I re-write this way: (10^x/9)-y/9 since they will have comment denominator?

Is that case, since x and y are positive, wouldn't the stem of the question itself denotes that it is not divisible by 9?

Yes, you can rewrite $$\frac{10^x-y}{9}$$ as $$\frac{10^x}{9}-\frac{y}{9}$$. But how can you say that $$\frac{10^x}{9}-\frac{y}{9}$$ will not be an integer?

For example, if x = 1 and y = 1, then $$\frac{10^x}{9}-\frac{y}{9}=1$$
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04 Sep 2017, 08:30
Does this rule hold true for any integer?

Quote:
(something not divisible by 3) - (something divisible by 3) = (something not divisible by 3)

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04 Sep 2017, 08:49
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Galiya wrote:
Does this rule hold true for any integer?

Quote:
(something not divisible by 3) - (something divisible by 3) = (something not divisible by 3)

Yes.

WHEN THE SUM OR THE DIFFERENCE OF NUMBERS IS A MULTIPLE OF AN INTEGER
1. 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.

2. 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.

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.

For more on this check:

5. Divisibility/Multiples/Factors

6. Remainders

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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04 Sep 2017, 08:59
Galiya wrote:
Does this rule hold true for any integer?

Quote:
(something not divisible by 3) - (something divisible by 3) = (something not divisible by 3)

Yeah this always holds true!

Sent from my iPhone using GMAT Club Forum mobile app

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22 Sep 2017, 06:41
I am struggling a bit to understand option B. How are we approaching it?

Statement (2) by itself is insufficient. If $$y = 1$$, the answer is "yes"; if $$y = 2$$, the answer is "no".

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22 Sep 2017, 07:25
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Shiprasingh1100 wrote:
I am struggling a bit to understand option B. How are we approaching it?

Statement (2) by itself is insufficient. If $$y = 1$$, the answer is "yes"; if $$y = 2$$, the answer is "no".

If y = 1, then 10^x + 1 could be 11, 101, 1001, ... so as given in the second statement, not divisible by 9. In this case, 10^x - 1 will be 9, 99, 999, ... so divisible by 9. Answer YES.

If y = 2, then 10^x + 1 could be 12, 102, 1002, ... so as given in the second statement, not divisible by 9. In this case, 10^x - 1 will be 8, 98, 998, ... so NOT divisible by 9. Answer NO.
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22 Sep 2017, 09:22
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Shiprasingh1100
It is stated in the question that x is positive integer i.e. x can be 1,2,3,4,5,6, and so on .So, 10^x will have have respective value 10,100,1000 and so on.
Thus in 2. when y=1 9,99,999, and so on will always be divisible by 9 whereas when y=2 it will not always be divisible by 9.

Hope you get it .
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22 Sep 2017, 20:00
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Shiprasingh1100 wrote:
I am struggling a bit to understand option B. How are we approaching it?

Statement (2) by itself is insufficient. If $$y = 1$$, the answer is "yes"; if $$y = 2$$, the answer is "no".

The question is asking if 10^x-y is divisible by 9. Since, for x=y=1; The answer is Yes

and for x=2, y=9; The answer is No.

Hence the option is not by itself sufficient.

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25 Sep 2017, 10:39
oirfan wrote:
How is statement 1 sufficient?

If we pick numbers for statement 1 we can get two different answers.
if x=2 and y=3, then 100-3=97, Answer NO
if x=2 and y=9, then 100-9=81, Answer YES

Am I missing something?

Even I make such mistakes, be careful. How should one avoid such mistakes?

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Re: M20-08   [#permalink] 25 Sep 2017, 10:39
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