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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
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Re: PS: Decimals [#permalink]
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



Please explain your answer


Walker u r ridiculous! :P 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.
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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
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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)
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[#permalink]
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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source [#permalink]
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?
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Re: PS: Decimals [#permalink]
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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



Please explain your 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
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Re: source [#permalink]
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. :)
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Re: Let D be a recurring decimal of the form D=0.a1 a2 a1 a2.... [#permalink]
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 .
so Answer is (C)
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Let D be a recurring decimal of the form D=0.a1 a2 a1 a2.... [#permalink]
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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