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# Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....

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Director
Joined: 21 Jul 2006
Posts: 957
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....  [#permalink]

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Updated on: 17 Aug 2013, 01:51
5
22
00:00

Difficulty:

65% (hard)

Question Stats:

57% (02:00) correct 43% (02:25) wrong based on 280 sessions

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Let D be a recurring decimal of the form D=0.a1 a2 a1 a2...., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18
b) 108
c) 198
d) 288
e) 158

Originally posted by tarek99 on 16 Dec 2007, 01:28.
Last edited by Bunuel on 17 Aug 2013, 01:51, edited 2 times in total.
Intern
Joined: 29 Jan 2013
Posts: 28
Re: Let D be a recurring decimal of the form D=0.a1 a2 a1  [#permalink]

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16 Aug 2013, 18:05
8
5
its straight question

give D=0.abababab (say - 1)
Multiply both sides by 100 i.e

100D = ab.abababab (say - 2)

now subtract 1 from 2 . That gives

99D = ab => D = ab/99... hence it should be multiplied by 99k to get an integer ab.
##### General Discussion
CEO
Joined: 17 Nov 2007
Posts: 2917
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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16 Dec 2007, 01:42
2
C

D=0.a1 a2 a1 a2....= 0.a1 a2 *(1+0.01+0.01^2+....+0.01^n+.....)

D= 0.a1 a2 * (1/(1-0.01))=0.a1 a2 /0.99=(a1 a2)/99

so, integer should be k*99. therefore, C 198=2*99
SVP
Joined: 29 Mar 2007
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16 Dec 2007, 22:45
tarek99 wrote:
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2...., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18
b) 108
c) 198
d) 288
e) 158

Walker u r ridiculous! so good at these.

Anyway, i remembered that 90/99=.90909090909090....

so one of the integers must be divisble by 9 and 11.

So we have d/99

198d/99 --> 2d So its C.
Manager
Joined: 03 Sep 2006
Posts: 161

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16 Dec 2007, 23:57
Guys, you are great!
But could somebody explain please how one follows from another:

(1+0.01+0.01^2+....+0.01^n+.....) = (1/(1-0.01)) ??

Thanks
CEO
Joined: 17 Nov 2007
Posts: 2917
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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17 Dec 2007, 00:47
Whatever wrote:
Guys, you are great!
But could somebody explain please how one follows from another:

(1+0.01+0.01^2+....+0.01^n+.....) = (1/(1-0.01)) ??

Thanks

∑q^i=1+q+q^2+...+q^n+...... - is a geometric sequence

for a finite geometric sequence: ∑q^i=(1-q^n)/(1-q)

for an infinite geometric sequence with 0<q<1:
q^n will goes to zero when n goes to +∞, so ∑q^i=(1-q^n)/(1-q)=1/(1-q)
Manager
Joined: 03 Sep 2006
Posts: 161

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17 Dec 2007, 01:00
walker wrote:
Whatever wrote:
Guys, you are great!
But could somebody explain please how one follows from another:

(1+0.01+0.01^2+....+0.01^n+.....) = (1/(1-0.01)) ??

Thanks

∑q^i=1+q+q^2+...+q^n+...... - is a geometric sequence

for a finite geometric sequence: ∑q^i=(1-q^n)/(1-q)

for an infinite geometric sequence with 0<q<1:
q^n will goes to zero when n goes to +∞, so ∑q^i=(1-q^n)/(1-q)=1/(1-q)

Thanks a lot.
I noticed that there is a sequence, but can't remember the formula of geometric sequence =)
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Joined: 12 Nov 2007
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27 Dec 2007, 07:18
I, too, would be very curious in knowning the source of this question.
Does someone has idea if the real GMAT could give questions of a similar difficulty?
It seemes VERY STRANGE to me that GMAT would want you to remember that formula of the summatory.

Bye,
Gianluca

bmwhype2 wrote:
what is the source of this question?
Manager
Joined: 21 May 2007
Posts: 103

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27 Dec 2007, 07:36
1
2
tarek99 wrote:
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2...., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18
b) 108
c) 198
d) 288
e) 158

For those who dont want to be troubled by the Geometric Series and its formula, here is a short-cut to remember:
any recurring decimal with 'n' recurring digits can be written as: A/B
A = the n recurring digits
B = 10^n - 1

Eg: 0.658658658..... = 658/999
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Joined: 16 Aug 2013
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16 Aug 2013, 09:19
gindro79 wrote:
I, too, would be very curious in knowning the source of this question.
Does someone has idea if the real GMAT could give questions of a similar difficulty?
It seemes VERY STRANGE to me that GMAT would want you to remember that formula of the summatory.

Bye,
Gianluca

bmwhype2 wrote:
what is the source of this question?

This question is taken from CAT 2000 paper.
Manager
Joined: 21 Aug 2012
Posts: 97
Re: Let D be a recurring decimal of the form D=0.a1 a2 a1  [#permalink]

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17 Aug 2013, 00:53
5
Two ways to solve this question:

1) When there is a recurring number i.e non-terminating number, the number can be expressed in the below form

Suppose N = 0.abababababab......

N can be written as
===> __
==>0.ab

To convert any recurring number to fraction form, divide the number by 9,99,999,9999, depending on the no. of recurring digits.

Here we have two recurring digits, so N=ab/99,
Similarly if we have,
====> _____
M = 0.abcde, them M = abcde/99999

NOTE: this is only for numbers in the form of 0.abcdef..... , not for normal recurring number 123123123....

Now in this question, we have
===> ____
D= 0.a1a2

Therefore, D=a1a2/99

Lets try each option and multiple it by D

For C, we have (a1a2/99)*198 => 2* a1a2, this will be an integer.

So option C is correct, for other option we are getting fractional terms.

2) Second way

D= 0.a1a2a1a2a1a2.......
===> ____
D=0.a1a2

Now,
=========> ____
100D= a1a2.a1a2

Now, 100D - D => a1a2 ( Both recurring parts will be subtracted)
99D = a1a2
D = a1a2/99

Rest same as explained 1st way.

Thanks,
Jai

KUDOS if it Helped..!!!
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MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636
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Senior Manager
Joined: 10 Jul 2013
Posts: 273
Re: Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....  [#permalink]

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17 Aug 2013, 10:29
tarek99 wrote:
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2...., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18
b) 108
c) 198
d) 288
e) 158

0.123412341234.................... = 1234/9999

0.34343434........................... = 34/99

0.543543543543..................... = 543/999

so, 0.a1a2a1a2a1a2a1a2................ = a1a2/99 and 198 is a multiple of 99 .
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Location: Azerbaijan
GMAT 1: 530 Q42 V21
GMAT 2: 600 Q42 V31
GMAT 3: 700 Q47 V38
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....  [#permalink]

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28 Sep 2017, 10:55
tarek99 wrote:
Let D be a recurring decimal of the form D=0.a1 a2 a1 a2...., where digits a1 and a2 lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18
b) 108
c) 198
d) 288
e) 158

I was not familiar with rule that if 0.12121212 then you can write it as 12/99. This is something I learned today.

But I answered the question correctly, with my own method. I just picked numbers. It could not be 1/3, because in this case only one number recurs (0.33333...). For the same reason it can not be 1/6 or 2/3. I tried 1/7, which revealed random pattern (0.142...). I tried 3/7, again random pattern.

Then I tried 3/11 - bum! this gave me 0.27272.... therefore I new that to produce an integer the number in the question must be multiplied by a multiple of 11. The only multiple of 11 in the answer choices was 198.
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Re: Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....  [#permalink]

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03 Oct 2018, 05:12
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Re: Let D be a recurring decimal of the form D=0.a1 a2 a1 a2....   [#permalink] 03 Oct 2018, 05:12