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

Question

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

Please explain your answer

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

adityapagadala · Accepted Answer

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.

jaituteja · Answer

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..!!!

walker · Answer

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

GMATBLACKBELT · Answer

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.

Whatever · Answer

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

walker · Answer

∑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)

Whatever · Answer

Thanks a lot. 
I noticed that there is a sequence, but can't remember the formula of geometric sequence =)

gindro79 · Answer

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

parsifal · Answer

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

astronaut91 · Answer

This question is taken from CAT 2000 paper.

Asifpirlo · Answer

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

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

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

so, 0.a1a2a1a2a1a2a1a2................ = a1a2/99 and 198 is a multiple of 99 .
so Answer is (C)

kablayi · Answer

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.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards