Problem:
Suppose
2+1+2+3+x211=53144
What is the value of x ?
Answer Choices:
A. 43
B. 87
C. 1514
D. 3837
E. 5352
Solution:
Answer (A): Subtracting 2 from both sides and taking reciprocals gives
1+2+3+x21=3853.
Subtracting 1 from both sides and taking reciprocals gives
2+3+x2=1538.
Subtracting 2 from both sides and taking reciprocals gives
23+x=815.
Thus x=2⋅815−3=415−412=43.
Note: The number e can be represented by the following continued fraction.
2+1+2+3+4+5+⋯43211
The fraction 53144≈2.717 is an approximation for e≈2.718.
The problems on this page are the property of the MAA's American Mathematics Competitions