Problem:
For a certain positive integer n less than 1000 , the decimal equivalent of n1 is 0.abcdef, a repeating decimal of period 6 , and the decimal equivalent of n+61 is 0.wxyz, a repeating decimal of period 4 . In which interval does n lie?
Answer Choices:
A. [1,200]
B. [201,400]
C. [401,600]
D. [601,800]
E. [801,999]
Solution:
Because n1=999999abcdef, it follows that n is a divisor of 106−1= (103−1)(103+1)=33⋅7⋅11⋅13⋅37. Because n+61=9999wxyz, it follows that n+6 divides 104−1=32⋅11⋅101. However, n+6 does not divide 102−1=32⋅11, because otherwise the decimal representation of n+61 would have period 1 or 2 . Thus n=101k−6, where k=1,3,9,11,33, or 99 . Because n<1000, the only possible values of k are 1,3 , and 9 , and the corresponding values of n are 95 , 297 , and 903. Of these, only 297=33⋅11 divides 106−1. Thus n∈[201,400]. It may be checked that 2971=0.003367 and 3031=0.0033.
The problems on this page are the property of the MAA's American Mathematics Competitions