soumya1989 wrote:
x = 1/3 + 1/9 + 1/27 + 1/37
| Column A | Column B |
| 101st digit after the decimal point in the decimal representation of x | 5 |
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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68% (02:19) correct
