Download Problems of the Month September 2014 Solutions

Problems of the Month
September 2014
Solutions
(1) Thirteen Powers of 2
Consider the set {3, 9, 81}. We can form the following consecutive powers of 3 by
way of multiplication where each consecutive power of 3 uses individual numbers in
the set at most once: 3, 9, 27, 81, 243, 729, 2187. Here 3 = 3, 9 = 9, 27 = (3)(9),
81 = 81, 729 = (9)(81), and 2187 = (3)(9)(81).
Now by way of multiplication and/or division, what is the smallest set of numbers
(give the set) such that each of the thirteen powers of 2 from 21 up to and including
213 can be obtained using each number in the set at most once per power? To be
clear, no element from the set you give should be used more than once to represent
any power of 2 in the thirteen powers of 2. Show your work and justify your answer.
Solution:
Every integer from 1 to 13 can be obtained by addition and subtraction
from the set {1, 3, 9} using each number at most one per result. Thus, the
corresponding powers of 2 can be obtained by multiplication and division
from the set {2, 8, 512} = {21 , 23 , 29 }.
(2) Maximum Area
Find the maximum area of a rectangle inscribed in a semicircle of radius r.
Solution:
Inscribing a rectangle in a semicircle positioned in quadrants I and II
with radius r results in the graph:
The area A of the rectangle inscribed is given by
A = 2xy
√
= 2x r2 − x2 .
Finding A0 (x) and setting it equal to zero yields
r
x = ±√ .
2
Using the positive result, we find that
r
A √
= r2 .
2
Thus, the maximum area of a rectangle inscribed in a semicircle of radius
r is r2 .
Congratulations to last month’s winner, Dylan Shields!
Also submitting a correct answer was Simon Peralta.