Problem:
What is the value of:
31​⋅42​⋅53​⋯2018​⋅2119​⋅2220​?
Answer Choices:
A. 4621​
B. 2311​
C. 1321​
D. 2132​
E. 221​
Solution:
Since every integer from 3 to 20 occurs once as a denominator and once as a numerator, they cancel each other out. After canceling every number out, we have only 1 and 2 left as numerators and 21 and 22 left as denominators. The remaining fraction is 21⋅221⋅2​. This simplifies to 4622​=2311​.
Answer: B​.
The problems on this page are the property of the MAA's American Mathematics Competitions