It has been a few years since I reviewed the criteria used to select solutions for publication. Let me do so again. As responses arrive, they are simply put together in neat
piles, with no regard to their date of arrival. When it is time for
me to write the column in which solutions are to appear, I first
weed out erroneous and illegible responses. For difficult problems, this may be enough; the most publishable solution becomes
obvious. Usually, however, many responses still remain. I try to
select a solution that supplies an appropriate amount of detail
and includes a minimal number of characters that are hard to set
in type. A particularly elegant solution is, of course, preferred,
as are contributions from correspondents whose solutions have
not previously appeared. I also favor solutions that are neatly
written, typed, or (especially) sent by e-mail, since these simplify typesetting.
M/A 1. Our wink-meister Rocco Giovanniello has a new problem
with a twist: this time two winks are initially missing. Specifically,
consider the following 5× 6 rectangle mostly filled with winks, as
shown. Each move is a horizontal or vertical jump of one wink
over another, landing on an empty square. The jumped-over wink
is removed. The goal is to remove all the winks save two, which
are in the initially empty two squares.
M/A 2. Another “logical hat problem” from Dick Hess.
Consecutive integers are chosen from 1, 2, 3, ... One is written
M/A 3. Our final regular problem is a generalization by Larry
on the hat of logician A and the other is written on the hat of
logician B. Each logician sees the other’s hat but not his or her
own. Each is error-free in reasoning and knows the situation.
They say in turn
A1: “I don’t know my number.”
B1: “I don’t know my number.”
A2: “I now know my number.”
What numbers are possible for A and B?
Stabile of problem 2 from the September/October 2016 issue.
Given an n-sided regular polygon (shown here for n = 5) with
the radius of the inscribed circle equal to R, draw radii to two adjacent points of tangency between the polygon and the inscribed
circle. Find the area of the shaded region as given by the boundary formed by the edges of the polygon and parallel lines from
the point P intersecting the edges, over the domain θ ε [0, 2π/n].
Can you determine what Sorab Vatcha’s favorite number is? It
has the following properties.
1. It is the smallest number equal to the sum of two positive
cubes in two ways.
2. It is the product of a palindrome pair.
3. Its three prime factors are the first, third, and fifth entries
in a sequence of five consecutive prime numbers.
4. The sum of its digits equals its largest prime factor.
N/D 1. Rocco Giovanniello offers a six-row triangular grid of 20
winks, with the apex (row 1) empty and rows 2 through 6 filled,
to be solved in 19 diagonal or horizontal jumps with the jumped-over wink removed and the last wink at the apex.
Greg Muldowney sent us the following solution, including
“action shots” of the wink moves.
In the solution diagrammed below, jumps 1 through 3 free up
space at the symmetry line, and all jumps thereafter are mirror-image pairs. Jumps 4/5 and 6/7 open symmetric positions in rows
3 and 5, enabling full vacation of the lower corners in jumps 8/9
and 10/11. Jumps 12/13 and 14/15 empty rows 5 and 6 and refill
row 2. This sets up a “home run” (jumps 16 through 19) in which
the apex wink traverses the four sides of a diamond and returns
to its starting point. Thus the first wink moved is also the last.