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M20-08

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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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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".


Answer: A
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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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New post 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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New post 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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Re: M20-08 [#permalink]

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New post 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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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



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New post 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!


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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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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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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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Re: M20-08 [#permalink]

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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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Re: M20-08 [#permalink]

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New post 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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