HP Forum Archive 21

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An interesting riddle Message #1 Posted by Don Shepherd on 13 Mar 2012, 9:37 p.m. Refer to this picture. It is a little figurine of a cardinal (tail-less, unfortunately, due to an accident) sitting on a feeding tray that doubles as a calendar. You notice the bar with the month names on the left, and you figure that there must be three of them, and you're right. You also notice the two visible dice-like cubes representing 13, March 13, today. The riddle: how many cubes are necessary to represent any possible date? Four days ago, it would require two cubes showing 09. You can see two day cubes; how many more are behind them?

Re: An interesting riddle Message #2 Posted by Paul Dale on 13 Mar 2012, 9:50 p.m., in response to message #1 by Don Shepherd I believe it can be done with just two cubes. Each cube needs the digits 0, 1 and 2. The remaining digits are spread over the rest of the faces and some rotational symmetry is taken advantage of. - Pauli

Re: An interesting riddle Message #3 Posted by Howard Owen on 13 Mar 2012, 10:03 p.m., in response to message #1 by Don Shepherd I get the same answer. The dice could be as follows: 0:1:2:3:4:5 and 0:1:2:7:8:9.

Re: An interesting riddle Message #4 Posted by Matt Agajanian on 13 Mar 2012, 10:13 p.m., in response to message #3 by Howard Owen Hold on! What happened to 6? Shouldn't a face of the second die be 6 instead of 0? Edited: 13 Mar 2012, 10:19 p.m. after one or more responses were posted

Re: An interesting riddle Message #5 Posted by Katie Wasserman on 13 Mar 2012, 10:19 p.m., in response to message #4 by Matt Agajanian Turn the 9 upside down to get the 6. You don't need a 3 on the other cube. Edited: 13 Mar 2012, 10:20 p.m. after one or more responses were posted

Re: An interesting riddle Message #6 Posted by Matt Agajanian on 13 Mar 2012, 10:20 p.m., in response to message #5 by Katie Wasserman Oops, silly me.

Re: An interesting riddle Message #7 Posted by M. Joury on 13 Mar 2012, 10:19 p.m., in response to message #4 by Matt Agajanian 6 / 9. See Pauli's note about "rotational symmetry". You have to have 0, 1, 2, on both cubes. 3 - 9 are only needed on one cube. 9 flipped is a 6. Cheers, -Marwan

Re: An interesting riddle Message #8 Posted by Matt Agajanian on 13 Mar 2012, 10:20 p.m., in response to message #7 by M. Joury Again, oops, silly me.

Re: An interesting riddle Message #9 Posted by M. Joury on 13 Mar 2012, 10:24 p.m., in response to message #8 by Matt Agajanian Katie beat me by seconds <g>. Only reason she got there first was that she typed less! She probably also types faster... -Marwan

Re: An interesting riddle Message #10 Posted by Katie Wasserman on 15 Mar 2012, 3:09 p.m., in response to message #9 by M. Joury I sometimes type fast, but make a zillion typos when I do. That's why you often see my edits shortly after I post something.

Re: An interesting riddle Message #11 Posted by M. Joury on 15 Mar 2012, 4:19 p.m., in response to message #10 by Katie Wasserman LOL. I also seem to have that particular disease. Cheers, -Marwan

Re: An interesting riddle Message #12 Posted by M. Joury on 13 Mar 2012, 10:17 p.m., in response to message #1 by Don Shepherd I also make it 2. Clever little riddle.

Re: An interesting riddle Message #13 Posted by Don Shepherd on 13 Mar 2012, 10:48 p.m., in response to message #1 by Don Shepherd You people are sooooo good. When I was eleven years old, my buddies and I thought it was so cool that the year on the US penny minted that year was the same if you looked at it upside-down. When was I born? What is the next year that will be the same upside-down?

Re: An interesting riddle Message #14 Posted by Jeff Kearns on 13 Mar 2012, 11:18 p.m., in response to message #13 by Don Shepherd 1961. 6119 Or.... after reading the next post, 1950 and 6009! Patience Grasshopper. Edited: 13 Mar 2012, 11:32 p.m.

Re: An interesting riddle Message #15 Posted by M. Joury on 13 Mar 2012, 11:18 p.m., in response to message #13 by Don Shepherd 1961 / 6009 Actually you asked when you were born -- that would be 1950. Edited: 13 Mar 2012, 11:19 p.m.

Re: An interesting riddle Message #16 Posted by Don Shepherd on 13 Mar 2012, 11:54 p.m., in response to message #15 by M. Joury Right, as usual.

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