perplexus.info :: Logic : Calendar Cubes pt 3

Calendar Cubes pt 3 (Posted on 2003-04-06)

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.

Submitted by Alan

Rating: 2.3000 (10 votes)

Solution: (Hide)

First lets start with the months. Although it may seem as though you may only have 6 it is possible to have more. Allow me to explain. First we could have the letter J for july and A for august. Well if on the dice they're both shown by having the dice on a diagonal you could have JA represent january. So that means that letter combination JA can represent july, august and January. Also letter combination MA can represent March, April, May. so on four faces of the cube it'd be written across JAMA. At which point the remaining two faces can have any other two months on them, but for sakes of the solution they need to be september and december(explained later) This gives us 8 months.
Next we see how many days can be represented. once again combination 12 could represent 1,2 and 12. On top of that depending how you write these numbers a 21 can also be represented. Follow that two up with a 3 and you've got 3 and 23 represented. Then follow that up with a 0 to create two more numbers (since it is now in a ring). This makes 30 and 10(upside down 1 and 0). Next we use a number 6 on one face to represent 6 and 9. Lastly any other number on the last face except 5. This gives us 10 digits(excluding 30 since not all months have it.) 8x10=80, Now we add 9 because only 9 of the months on the cube have 30 days.(this is why we needed september and december) Thus, giving us a grand total of 89 days. But one more day can be represented using the zero on the number cube and the s(for september) on the other cube we can make the zero the letter O for october and the S of september as a 5, thus giving us one more day.
Making the grand total 90 days.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
3 more dates than the given solution	Dej Mar	2010-07-03 14:22:11
all twelve months	Cory Taylor	2003-04-14 04:22:30
Posted solution does not work	fwaff	2003-04-14 03:38:29
re: What?	fwaff	2003-04-10 22:52:35
re: low rating	fwaff	2003-04-10 22:24:50
low rating	benard	2003-04-10 12:11:25
What?	Alan	2003-04-10 10:38:22
re(5): Attempt at a solution - 2 more combinations	fwaff	2003-04-08 22:12:44
re(4): Attempt at a solution - 2 more combinations	Charlie	2003-04-08 15:15:40
re(3): Attempt at a solution - 2 more combinations	Alan	2003-04-08 10:13:22
re(2): Attempt at a solution - 2 more combinations	fwaff	2003-04-07 03:34:09
re: Attempt at a solution - more combinations	fwaff	2003-04-06 22:48:14
Attempt at a solution (probably too naive)	Charlie	2003-04-06 16:28:35
Different blocks?	Gamer	2003-04-06 16:07:33

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