Problem:
The repeating decimals 0.ababab and 0.abcabcabc satisfy
0.ababab+0.abcabcabc=3733
where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.
Solution:
Let x=0.ababab and y=0.abcabcabc. Then x=99ab and y=999abc, so
3733=99ab+999abc=33⋅11⋅37111ab+11abc
Because this fraction must reduce to 3733, it must be the case that the numerator 111ab+11abc is a multiple of 11. Thus 111ab must be a multiple of 11, and because 111 is relatively prime to 11, it follows that ab is a multiple of 11 and a=b. Thus
33⋅3727⋅33=3733=33⋅11⋅37111aa+11aac=33⋅37111a+aac=33⋅37221a+c
Thus, 891=33⋅27=221a+c, and it follows that a=4 and c=7. Therefore the three-digit number abc is 447.
The problems on this page are the property of the MAA's American Mathematics Competitions