List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..

Question

List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..., 0.989898.... If each value in list L is expressed as a fraction in lowest terms, what is the sum of the denominators of all 98 fractions?

(A) 5,096
(B) 5,608
(C) 6,770
(D) 6,894
(E) 9,702

(A) 5,096
(B) 5,608
(C) 6,770
(D) 6,894
(E) 9,702

GMATinsight · Answer

0.010101... = \frac{01}{99} 
 0.020202... = \frac{02}{99} 
 0.030303... = \frac{03}{99} 
----
----
----
 0.989898.... = \frac{98}{99}

These are 98 terms
But factors of 99 are {1, 3, 9, 11, 33, 99}

i.e. in Lowest form 
 \frac{3}{99 }= \frac{1}{33} 
 \frac{9}{99 }= \frac{1}{11} 
 \frac{11}{99 }= \frac{1}{9} 
 \frac{33}{99 }= \frac{1}{3}

i.e. 98-4 = 94 terms would have denominator 99 and 4 other terms would have denominators as {33, 11, 9, 3}

Sum of all 98 denominators = 99*94+(33+11+9+3) = 9362

ANswer: Option E

Bunuel  : Please check options. I hope I am not making any mistake

DeeRocket · Answer

GMATinsight

I just want to extend your approach-

For factor 33, 33/99 and 66/99 will be reduced to least fraction as 1/3 and 2/3, respectively. So, for both of them 3 will be added i.e. 2*3 = 6
For factor 11, 11/99, 22/99, ...88/99 will be reduced to least fraction as 1/9, 2/9, ... 8/9 (here we will ignore 33 and 66 as they are covered in 33), respectively. So, for all remaining 6 values 9 will be added i.e. 6*9 = 54
Similarly, for factor 9 we will get denominator as 11 for 10 values i.e. 10*11 = 110 and for factor 3 we will get denominator as 33 for 20 values (ignoring repetitions) i.e. 20*33 = 660

Final answer will be : 98*60 + 6 +54 +110 + 660 = 6770

Ans. C

Correct me if I have made some mistake.

Adarsh_24 · Answer

Deerocket has explained it well.

It’s important to check for repetion.

We can start from 33 - it has only 2 points. 33 and 66.
Then we go to 11- since 11 is a factor of 33, we have overlap here. So need to remove 33 and 66 already considered.
Then to 9- 9 is not factor of 33 or 11. So no need to remove any.
Then to 3- 3 is factor of both 9 and 33. Between 1 and 98 inclusive we have 32 multiples of 3.
33 had 2 multiples. 9 has 10 multiples. So 32-12=20

Kinshook · Answer

List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..., 0.989898....

If each value in list L is expressed as a fraction in lowest terms, what is the sum of the denominators of all 98 fractions?

Let x = .010101....
100x = 1.010101....
99x = 1
x = 1/99

List L = 1/99, 2/99, 3/99=1/33, 4/99, 5/99, 6/99 = 2/33, 7/99, 8/99, 9/99 = 1/11, 10/99, 11/99 = 1/9, ... 98/99

If numerators are multiples of factors of 99, denominator is reduced by common factor of 99 & numerator.

99 = 3^2*11

Factors of 99 = 3, 9, 11, 33, 99
Multiples of 33 = 33, 66; Denominator = 3 in fraction in lowest terms instead of 99; Number of such terms = 2
Multiples of 11 but not 33 = 11, 22, 33, 44, 55, 66, 77, 88; Denominator = 9 in fraction in lowest terms instead of 99; Number of such terms = (88-11)/11 + 1 - 2 = 8 - 2 = 6
Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90: Denominator = 11 in fraction in lowest terms instead of 99; Number of such terms = (90-9)/9 + 1 = 10
Multiples of 3 but not 9 or 33 = 3, 6, 12, 15, ...., 96; Denominator = 33 in fraction in lowest terms instead of 99; Number of such terms = (96 - 3)/3 + 1 - 10 = 32 - 10 - 2 = 20

The sum of the denominators of all 98 fractions = 99*98 - 20*(99-33) - 10*(99-11) - 6*(99-9) - 2*(99-3) = 6770

IMO C

List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..

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List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..

Prep Toolkit

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