It is currently 22 Mar 2018, 16:33

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M20-08

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

16 Sep 2014, 01:08
Expert's post
13
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

55% (01:12) correct 45% (01:05) wrong based on 159 sessions

HideShow timer Statistics

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

16 Sep 2014, 01:08
2
KUDOS
Expert's post
2
This post was
BOOKMARKED
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".

_________________
Intern
Joined: 15 Nov 2015
Posts: 2

Show Tags

28 Mar 2016, 21:46
1
KUDOS
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?
Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

28 Mar 2016, 22:16
1
KUDOS
Expert's post
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.
_________________
Intern
Joined: 10 May 2017
Posts: 27

Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

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 $$\frac{10^x}{9}-\frac{y}{9}=1$$
_________________
Manager
Joined: 16 Jan 2011
Posts: 107

Show Tags

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)
Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

04 Sep 2017, 08:49
Expert's post
1
This post was
BOOKMARKED
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 ! ! !
_________________
Intern
Joined: 04 Sep 2017
Posts: 1

Show Tags

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
Intern
Joined: 28 Mar 2017
Posts: 4

Show Tags

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".
Math Expert
Joined: 02 Sep 2009
Posts: 44400

Show Tags

22 Sep 2017, 07:25
1
KUDOS
Expert's post
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.
_________________
Intern
Status: Don't watch the clock,Do what it does, Keep Going.
Joined: 10 Jan 2017
Posts: 46

Show Tags

22 Sep 2017, 09:22
2
KUDOS
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
Manager
Joined: 21 Jul 2017
Posts: 96
Location: India
GMAT 1: 650 Q47 V33
GPA: 4

Show Tags

22 Sep 2017, 20:00
1
KUDOS
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.
Manager
Joined: 04 Jul 2017
Posts: 61
Location: India
Concentration: Marketing, General Management
GPA: 1
WE: Analyst (Consulting)

Show Tags

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?
Re: M20-08   [#permalink] 25 Sep 2017, 10:39
Display posts from previous: Sort by

M20-08

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.