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M20-08

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Bunuel
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M20-08 [#permalink] New post   16 Sep 2014, 01:08
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If \(x\) and \(y\) are positive integers, is \(10^x - y\) divisible by 9 ?


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

(2) \(10^x + y\) is not divisible by 9
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Bunuel
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Re M20-08 [#permalink] New post   16 Sep 2014, 01:08
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Official Solution:


If \(x\) and \(y\) are positive integers, is \(10^x - y\) divisible by 9 ?

Note that for any positive integer \(x\), \(10^x\) will have a digit sum equal to 1, so it will not be divisible by 3.

(1) \(y\) is divisible by 3.

In this case, \(10^x - y\) = (something not divisible by 3) - (something divisible by 3) = (something not divisible by 3). Since \(10^x - y\) is not divisible by 3, it is also not divisible by 9. Sufficient.

(2) \(10^x + y\) is not divisible by 9.

If \(y = 1\), the digit sum of \(10^x + y\) will be 2, so it won't be divisible by 9, but \(10^x - y\) will have only 9s as its digits, so it will be divisible by 9. However, if \(y = 2\), the digit sum of \(10^x + y\) will be 3, so it won't be divisible by 9, but \(10^x - y\) will have some number of 9s followed by an 8, so it won't be divisible by 9 either. Not sufficient.


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Re: M20-08 [#permalink] New post   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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Re: M20-08 [#permalink] New post   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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Re: M20-08 [#permalink] New post   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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Re: M20-08 [#permalink] New post   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 multiples 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 the other is not, then their sum and difference will NOT be multiples 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 multiples 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



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Re: M20-08 [#permalink] 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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Re: M20-08 [#permalink] New post   23 May 2017, 06:34
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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] 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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Re: M20-08 [#permalink] New post   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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Re: M20-08 [#permalink] New post   23 Apr 2023, 13:22
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M20-08 [#permalink]
23 Apr 2023, 13:22
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