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".


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?

100 - 9 = 91, not 81.
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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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Does this rule hold true for any integer?

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

• Theory
  Divisibility and Remainders on the GMAT
  Tips and hints: Divisibility/Multiples/Factors
  3 Deadly Mistakes you must avoid in LCM-GCD Questions
  How Well Do You Know Your Factors?
  THE MOST EFFICIENT WAY TO STUDY LEAST COMMON MULTIPLES ON THE GMAT
  
  • Questions
    The E-GMAT Primes Trio - 3 Ques on No. of factors & Prime factors
    DS Questions
    PS Questions

6. Remainders

• Theory
  REMAINDERS
  Remainders: Tips and hints
  Divisibility and Remainders on the GMAT
  Remainders on the GMAT
  A Remainders Shortcut for the GMAT
  All About Negative Remainders on the GMAT
  A Tricky Question on Negative Remainders
  Compilation of tips and tricks to deal with remainders
  Cyclicity and the remainders on the GMAT
  GMAT Question Patterns: Remainders
  
  • Questions
    Units digits, exponents, remainders problems
    DS Questions
    PS Questions

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
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Yeah this always holds true!


Sent from my iPhone using GMAT Club Forum mobile app

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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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 .
Kudos if you like my reply

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.

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

If x=y=1

Then statement 1: 10 - 1 = 9. Then this is divisible by 9

How can y be 1 for the first statement? (1) says y is divisible by 3. Is 1 divisible by 3?
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As per the explanation provided, point 3 is applied in this case which says it may or may not be divisible by that number. So, how can we conclude below:
(something not divisible by 3) - (something divisible by 3) = (something not divisible by 3)

Please read carefully. Solution says, (something NOT divisible by 3) - (something divisible by 3) = (something not divisible by 3). It's what point 2 above says, NOT 3.
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My bad! thank you for pointing it out.