Problem:
Let a0​=2,a1​=5, and a2​=8, and for n>2 define an​ recursively to be the remainder when 4(an−1​+an−2​+an−3​) is divided by 11 . Find a2018​⋅a2020​⋅a2022​.
Solution:
Calculating the first few values of ak​ shows that they repeat with period 10 and a8​=7. It follows that a2018​⋅a2020​⋅a2022​=a8​⋅a0​⋅a2​=7⋅2⋅8=112​.
The problems on this page are the property of the MAA's American Mathematics Competitions