Problem:
Let S be the set of all rational numbers r,0<r<1, that have a repeating decimal expansion of the form
0.abcabcabc…=0.abc
where the digits a,b,c are not necessarily distinct. To write the elements of S as fractions in lowest terms, how many different numerators are required?
Solution:
First note that
0.abc=999abc​
and that 999=3337.
If abc is divisible by neither 3 nor 37, then this fraction is already in lowest terms. By the Inclusion-Exclusion Principle, there are
999−(3999​+37999​)+(3⋅37999​)=999(1−31​)(1−371​)=648
such numbers.
Some of the reduced fractions may have numerators that are divisible by 3 or 37. Such fractions must have the form
37k​, where k is a multiple of 3 but not a multiple of 37
or
3ml​, where l is a multiple of 37 but not a multiple of 3, and m=1,2,3
There are no fractions of the second type in S, since any fraction of this form is greater than 1. There are 12 fractions of the first type in S, one for each of k=3,6,…,36. Thus the number of distinct numerators in the set of reduced fractions is 648+12=660​.
The problems on this page are the property of the MAA's American Mathematics Competitions