Problem:
The fraction
9921​=0.bn−1​bn−2​…b2​b1​b0​​
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−1​bn−2​…b2​b1​b0​⋅bn−1​bn−2​…b2​b1​b0​​
Subtracting the original equation gives
99210n−1​=bn−1​bn−2​…b2​b1​b0​
Thus 10n−1=992⋅bn−1​bn−2​…b2​b1​b0​. 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−1​102k=(102−1)k=0∑N−1​ak​102k
and so
k=0∑N−1​102k+k=0∑N−1​ak​102k=k=1∑N​ak−1​102k
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−1​bn−2​…b2​b1​b0​=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