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101st digit after the decimal point
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soumya1989
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Joined: 16 May 2014
Posts: 597
GRE 1: Q165 V161
101st digit after the decimal point [#permalink] New post  08 May 2016, 03:00
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64% (01:53) correct 35% (01:20) wrong based on 34 sessions

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x = 1/3 + 1/9 + 1/27 + 1/37
Column AColumn B
101st digit after the decimal point in the decimal representation of x5


A bit hard one. Truth be told chances of getting a question like this in actual GRE is quite slim. Still good to practise these.
Kudos to the answer with the best explanation.
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Sonalika42
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Joined: 12 Jan 2016
Posts: 142
Re: 101st digit after the decimal point [#permalink] New post  26 May 2016, 08:18
B
bcoz 101st digit will be zero
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rapsjade
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Joined: 23 Jan 2016
Posts: 133
Re: 101st digit after the decimal point [#permalink] New post  26 May 2016, 19:42
1
the value is 508/999; This is a repeating decimal .508508.....; Hence 99th digit will be 8, so 101th digit will be 0. Hence answer is B
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AkashDU
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Joined: 16 Aug 2015
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Re: 101st digit after the decimal point [#permalink] New post  27 May 2016, 17:58
soumya1989 wrote:
x = 1/3 + 1/9 + 1/27 + 1/37
Column AColumn B
101st digit after the decimal point in the decimal representation of x5


A bit hard one. Truth be told chances of getting a question like this in actual GRE is quite slim. Still good to practise these.
Kudos to the answer with the best explanation.




LCM of 27 & 37 is 999. So whatever goes to numerator, will repeat itself in the decimal reprsentation. In this case it is 508.
Now, 101 leaves a remainder of 2 when it is divided by the cyclicity period of the representation. Here, the cyclicity period is 3.
Therefore the 2nd digit of 508 will be the 101th number.

Ans: B
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IshanGre
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Joined: 29 Nov 2017
Posts: 190
Location: United States
GRE 1: Q142 V146
WE:Information Technology (Computer Software)
Re: 101st digit after the decimal point [#permalink] New post  28 Apr 2018, 20:43
AkashDU wrote:
soumya1989 wrote:
x = 1/3 + 1/9 + 1/27 + 1/37
Column AColumn B
101st digit after the decimal point in the decimal representation of x5


A bit hard one. Truth be told chances of getting a question like this in actual GRE is quite slim. Still good to practise these.
Kudos to the answer with the best explanation.




LCM of 27 & 37 is 999. So whatever goes to numerator, will repeat itself in the decimal reprsentation. In this case it is 508.
Now, 101 leaves a remainder of 2 when it is divided by the cyclicity period of the representation. Here, the cyclicity period is 3.
Therefore the 2nd digit of 508 will be the 101th number.

Ans: B



How did you get 508 in numerator?
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