Problem Statement: Arrange the digits from 0 – 9 on two blank cubes, in order to be able to form a ‘date calendar’ which will be able to denote any given day of each month.
Solution/Discussion: The fun thing about this problem is that there’s one subtle thing you have to notice first, and then, after that, there’s a nice ‘brick wall’ in your way. To pin those down, I’ll let the e-mail exchange from Dan Felshin do the talking:
The first response from Dan went this way:
Larry,
Thanks for great article. I solved this with the aid of my wife. First I figured that a cube has six sides. Then we had to tell how many duplicates which are 11 and 22. Thank goodness there is no 33rd day. We then realized that there must me 0, 1, and 2 on each cube and the remaining six digits are split up.
I am 81 years old and I still do algebra problems from a book and play chess against the computer. Both are pure logic. Algebra is not really math, only logic. My wife does Soduko which is also logic, not arithmetic.
Dan Felshin, Springfield
Dan had noted the subtle thing mentioned above: You need a 0 on each cube! (You can’t put a 0 on one cube and fit all 9 other digits on the other!)
But Dan had miscounted. 🙂 After that, there are seven remaining digits to place, not six. (Which creates the brick wall: Seven digits on only six remaining faces – how can we do this?)
When I wrote Dan back, congratulating him, and pointing out the glitch, I get this very quick response:
I looked at it again and saw that it required one more number than would fit. And then it dawned on me: The 6 could also be used as a 9 if turned upside down
And that finishes the problem!! Good thinking, Dan (and wife)!! Keep up the good work!!
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