Problem:
The fraction
9921=0.bn−1bn−2…b2b1b0
where n is the length of the period of the repeating decimal expansion. What is the sum b0+b1+⋯+bn−1 ?
Answer Choices:
A. 874
B. 883
C. 887
D. 891
E. 892
Solution:
Note that
99210n=980110n=bn−1bn−2…b2b1b0⋅bn−1bn−2…b2b1b0
Subtracting the original equation gives
99210n−1=bn−1bn−2…b2b1b0
Thus 10n−1=992⋅bn−1bn−2…b2b1b0. It follows that 10n−1 is divisible by 11 and thus n is even, say n=2N. For 0≤j≤N−1, let aj=10b2j+1+b2j. Note that 0≤aj≤99, and because
102−1102N−1=1+102+104+⋯+102(N−1)
it follows that
k=0∑N−1102k=(102−1)k=0∑N−1ak102k
and so
k=0∑N−1102k+k=0∑N−1ak102k=k=1∑Nak−1102k
Considering each side of the equation as numbers written in base 100 , it follows that 1+a0≡0(mod100), so a0=99 and there is a carry for the 102 digit in the sum on the left side. Thus 1+(1+a1)≡a0=99(mod100) and so a1=97, and there is no carry for the 104 digit. Next, 1+a2≡a1=97 (mod100), and so a2=96 with no carry for the 106 digit. In the same way aj=98−j for 1≤j≤98. Then 1+a99≡a98=0(mod100) would yield a99=99, and then the period would start again. Therefore N=99 and bn−1bn−2…b2b1b0=0001020304…969799. By momentarily including 9 and 8 as two extra digits, the sum would be (0+1+2+⋯+9)⋅20=900, so the required sum is 900−9−8=883.
The problems on this page are the property of the MAA's American Mathematics Competitions