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This can be written as -

13/27 + 4/111 = 0.0360360360... + 0.481481.. = 0.517517

Hence, 7
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1/111+1/37=4/111=21st and 22nd digit=6,0
1/27+1/9+1/3=21st and 22nd digit 1,4
Answer=7(1+6)
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1/3 = 0.3333(non- terminating)

Thus, 21st digit is 3

1/9 = 0.11111(non-terminating)

Thus, 21st digit is 1

1/27 = 0. 037037

Thus number repeats in sequence of 3 digits - 037, thus 21st digit number is 7

1/37 = 0.027. Thus, same as above, 21st digit is 7

1/111 = 0.009, thus, for same reason as above, 21st digit is 9

Adding above, 3+1+7+7+9 = 7

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Expressing each of the fractions in the decimal form:

1/3 = 0.333333333333.... --> 21st digit is 3
1/9 = 0.111111111111... --> 21st digit is 1
1/27 = 0.037037037037...--> 21st digit is 7
1/37 = 0.027027027027.. --> 21st digit is 7
1/111= 0.009009009...... --> 21st digit is 9


9+7+7+1+3 = 27, 7 is the 21st digit and 2 is carried over for sum in 20th digit.

C
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gmatbusters

Official Solution:


\(\frac{1}{111} + \frac{1}{37} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3}\)
= \(\frac{9}{999} + \frac{27}{999} + \frac{37}{999} + \frac{111}{999} + \frac{333}{999}\)
= \(\frac{517}{999}\)
= 0.517517517...
It is clear that 7 at places 3rd, 6th, 9th... multiple of 3.
Hence 21st digit is 7

Answer is C

Method to convert fraction (with only 9 as digits in denominator) to decimal


  • If the fraction is abc/999, decimal will be 0.abcabcabc...
    We need to just repeat the numerator indefinitely after the decimal.
  • if the fraction is ABC/99999, decimal will be 0.00ABC00ABC
    We need to add zeros after zero at the start of numerator and repeat it indefinitely.

gmatbusters

Weekly Quant Quiz Question - 5



What is the 21st digit of the decimal \(\frac{1}{111} + \frac{1}{37} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3}\) ?
a) 5
b) 6
c) 7
d) 8
e) 9

gmatbusters

I guess there is a typo. The last fraction should be 333/999=1/3.

It could be confusing so if you can edit that would be great. Thank you!
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It is helpful to understand the following about repeating decimals:

When ONE digit is repeated, a denominator of 9 is implied:
\(0.\overline{5} = \frac{5}{9}\)
When TWO digits are repeated, a denominator of 99 is implied:
\(0.\overline{27} = \frac{27}{99} = \frac{3}{11}\)
When THREE digits are repeated, a denominator of 999 is implied:
\(0.\overline{006} = \frac{006}{999} = \frac{2}{333}\)


GMATBusters

Weekly Quant Quiz Question - 5



What is the 21st digit of the decimal \(\frac{1}{111} + \frac{1}{37} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3}\) ?
a) 5
b) 6
c) 7
d) 8
e) 9

Sum of the given fractions \(= \frac{517}{999}\)
As illustrated by the blue case above, a denominator of 999 implies repetition of the 3 digits in the numerator:
\(0.\overline{517}\)
Since every 3rd digit to the right of the decimal will be 7, the 21st digit = 7

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