All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Calendar Cubes pt 3 (Posted on 2003-04-06) Difficulty: 4 of 5
In the problem Calendar cubes you figured out the maximum amount of days you could fit on two cubes by putting numbers on both cubes and using the faces of the cubes to combine and make more numbers.
In Calendar cubes pt 2 You figured out how many months you could fit on two cubes.
Now in calendar cubes pt 3 you must figure out how many days of the year you can fit on two cubes. E.G one cube says mar(for march) and another says 5. so you could make the date march 5 and that would count as one date.

To represent months you may use
a) the first letter of that month
b) the first and second letter of that month
c) the first three letters of that months. So for january you could use either j, ja or jan to represent that month.
Also no two letters or letter combinations can represent the same month. So j cannot stand for june and july, but you can have j stand for june and ju stand for july. Also note that one month symbol (lets say au for august) can be on 1 face of 1 cube.

See The Solution Submitted by Alan    
Rating: 2.3000 (10 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(5): Attempt at a solution - 2 more combinations | Comment 7 of 14 |
(In reply to re(4): Attempt at a solution - 2 more combinations by Charlie)

What I was explaining badly is that on the 'month' cube I'm assuming S represents September and on the 'day' cube Z0 represents 20 (similarly 0Z represents 02 = 2). From these two faces it's possible to get three different dates:

S 0Z - 2nd of September
S Z0 - 20th of September (invert the 'day' cube)
NO S - 5th of November (rotate the 'day' cube clockwise through 90deg so that it shows NO reading vertically downwards and interchange the two cubes such that the 'month' cube is now showing the day and the 'day' cube is now showing the month) This relies on a fairly relaxed approach to font usage such that Z=2 and S=5.

Similarly I'm rotating the M used to represent March to become a 3 to give two faces that show M 0Z (2nd March), M Z0 (20th March) and NO 3 (3rd November).

I said it was tenuous in my previous comment, but I presume that the point of the problem is to find interchangeable/invertible numbers and letters to maximise the number of combinations.

Charlie: you're correct with Bonfire Night = "Guy Fawkes Day". 5th November is when we island monkeys celebrate the capture of our most notorious terrorist/freedom-fighter (depending on your perspective) Guido (Guy) Fawkes as he attempted to blow up the houses of parliament with three dozen barrels of gunpowder in 1605.
  Posted by fwaff on 2003-04-08 22:12:44

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (24)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information