bkpolymers1617 wrote:
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
STEP 1: CONVERT DECIMALS TO FRACTIONS
X= 0.010101..... ------- (I)
100X= 1.010101.... -------- (II)
(II)-(I) =>
99X = 1 OR X = 1/99
SIMILARLY, 2/99, 3/99.......... 98/99
Series: 1/99, 2/99, 3/99, 4/99...... 98/99
Number of Terms = 98
STEP 2: PRIME FACTORIZATION OF 99
99 = 3*3*11
only, 3 and 11 are factors
(Note (3 is a factor of 9 too)
STEP 3: BACK CALCULATION TO REACH ANSWER
I.
99*98 = (100-1)*98= 9800-98 =9702
THEREFORE, E OUT!!
II. A. 11/99, 22/99, 33/99, 44/99, 55/99, 66/99, 77/99, 88/99 (NOTE 33/99 and 66/99 will be used in 3s too, But i have used these two here in 11s)
1/9, 2/9,
3/9, 4/9, 5/9,
6/9, 7/9, 8/9
REDUCTION IN DENOMINATOR = 99-9 = 90
Number of terms involved = 8
total reduction from 9702= 90*=720
THEREFORE, 9702- 720= 8982
II. B. 3/9 and 6/9 above can be further reduced to 1/3 and 2/3
total reduction = 2*(9-3) =12
8982-12 = 8970
III. 3/99, 6/99, 9/99, 12/99, 15/99, .......... 96/99
3*1=3
3*33=99
and
3*32= 96
Number of factors involved = 32
NOTE : IN II. we have used two terms=> 33/66, 66/99
THEREFORE,
Number of factors remaining= 30
can be further reduced to
1/33, 2/33, 3/33, ....... , 32/33
REDUCTION = (99-33) =66
NUMBER OF TERMS = 30
TOTAL FURTHER REDUCTION = 66*30= 1980
THEREFORE,
8970-1980= 6990
IV. Now, above 30 factors that have been reduced to
1/33, 2/33, 3/33, ....... , 32/33
Out of the above
3/33, 6/33, 9/33, ..... 30/33
can further be reduced
Number of terms involved = 10
(??why?? as 3*1= 3; 3*10=30; 3*11=33(not possible))
TO
1/11, 2/11........ 10/11
REDUCTION = (33-11)= 22
Number of terms = 10
TOTAL FURTHER REDUCTION = 220
THEREFORE,
6990-220 =6770
ANSWER (C)
DROP KUDOS, IF THIS HELPS!!