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Topic: 90>100? (Read 530 times) |
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rmsgrey
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There is a device in my house which has a number of buttons - the numbers 0-9, an activation button and assorted mode selection buttons (which are irrelevant). The device is operated by entering a number of up to 4 digits using the buttons, optionally pressing a mode selection button then pressing the activation button.
Being lazy, I regularly make use of the device's belief that 90>100 to save a keypress. Given that the device believes that 10>9, 1000>900, x>=0 and 9999>=x for any arbitrary number x entered, can you explain the device's strange belief?
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aero_guy
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Re: 90>100?
« Reply #1 on: Dec 2nd, 2003, 9:03am » |
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I am pretty sure this is on here in a different form. I would find it, but every time I hit the search button it crashes my browser.
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william wu
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Re: 90>100?
« Reply #2 on: Dec 2nd, 2003, 9:26am » |
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When your browser crashes, does an error message pop up? Or does your browser just freeze? The search function takes a little while, on the order of 7 seconds or so.
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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aero_guy
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Re: 90>100?
« Reply #3 on: Dec 2nd, 2003, 3:46pm » |
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It crashes instantly. It just says that it has generated errors and will therefore close. I am running netscape 7.1.
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Icarus
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Re: 90>100?
« Reply #4 on: Dec 2nd, 2003, 4:27pm » |
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Those evil Sumerians: more than 4000 years gone, but still causing confusion. 
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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aero_guy
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Re: 90>100?
« Reply #5 on: Dec 3rd, 2003, 5:57am » |
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You could always just push the 100 button itself and save even more time, usually it does not require use of the activation button.
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Dudidu
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Re: 90>100?
« Reply #6 on: Dec 3rd, 2003, 7:37am » |
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rmsgrey hi,
Why can't I say (or maybe can I say) that the device works in a (modulo 7) field. Thus,
* 10>9 [to] 10 (mod 7) = 3 > 2 = 9 (mod 7)
* 1000>900 [to] 1000 (mod 7) = 6 > 4 = 900 (mod 7)
* x>=0 and 9999>=x for any arbitrary number x entered - practically, we even get x>=0 and 6>=x for any arbitrary number x entered.
* 90>100 [to] 90 (mod 7) = 6 > 2 = 100 (mod 7)
which satisfies all the conditions...
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Icarus
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Re: 90>100?
« Reply #7 on: Dec 3rd, 2003, 9:31am » |
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on Dec 3rd, 2003, 5:57am, aero_guy wrote:| You could always just push the 100 button itself and save even more time, usually it does not require use of the activation button. |
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But it's 90, not 100, that he wants.
Dudidu - that works, but I know of no common device that uses modulo 7 arithmetic. rmsgrey's device is quite common. I have one too, and regularly use the same trick.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Dudidu
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Re: 90>100?
« Reply #8 on: Dec 3rd, 2003, 9:51am » |
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on Dec 3rd, 2003, 9:31am, Icarus wrote:Dudidu - that works, but I know of no common device that uses modulo 7 arithmetic. rmsgrey's device is quite common. I have one too, and regularly use the same trick.
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So... if we're talking about a real device ( ) then it might be somekind of a microwave/oven since assuming that its operation time is expressed as MM:SS, all the conditions follow (specifically, 100<90 since 1 minute < 90 seconds).
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aero_guy
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Re: 90>100?
« Reply #9 on: Dec 3rd, 2003, 9:53am » |
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Actually he wants 100 and just types 90 to save typing an extra digit. And even if he did want 90 he could just hit the 100 button twice and end it after 90.
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DeMark
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Re: 90>100?
« Reply #10 on: Dec 3rd, 2003, 12:18pm » |
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Maybe write it from right to left 001<09? No? 
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Icarus
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Re: 90>100?
« Reply #11 on: Dec 3rd, 2003, 3:23pm » |
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on Dec 3rd, 2003, 9:53am, aero_guy wrote:| Actually he wants 100 and just types 90 to save typing an extra digit. And even if he did want 90 he could just hit the 100 button twice and end it after 90. |
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I can't speak for rmsgrey, but when I do this, it is 90 that I want. The extra keypress is saved from the more conventional way of entering the same information. And entering an extra keypress is much easier than standing around to catch it at 90.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Dudidu
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Re: 90>100?
« Reply #12 on: Dec 3rd, 2003, 10:52pm » |
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on Dec 3rd, 2003, 12:18pm, DeMark wrote:It seems incorrect since when we look at 10>9 (for example) and write it from right to left, we get 01 > 9 ( ) which is not true.
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| « Last Edit: Dec 3rd, 2003, 10:54pm by Dudidu » |
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rmsgrey
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Re: 90>100?
« Reply #13 on: Dec 4th, 2003, 7:19am » |
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Firstly, I have no 100 buttons on my device. Looking back at my original description, unless they fell under the heading of "assorted mode selection buttons (which are irrelevant)", they are excluded, as is a 90 button, and any similar variants.
Dudidu: 9999=7000+2800+140+56+3, so working mod 7, while ingenious, fails.
For the record, the correct answer is ::microwave:: in case anyone's still agonising over it.
Generally, when I enter 90, I want something "a bit more than" 100. Actually my most common use recently has been
::
to heat pies, for which I enter {6}{0}{full power}{2}{0}{0}{hold}{9}{0}{cook}
::
(not that I expect anyone to still want things hidden by this stage, but just in case...)
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Dudidu
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Re: 90>100?
« Reply #14 on: Dec 4th, 2003, 7:33am » |
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on Dec 4th, 2003, 7:19am, rmsgrey wrote:| Dudidu: 9999=7000+2800+140+56+3, so working mod 7, while ingenious, fails. |
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I understand that you were looking for a real device, but I do not understand () the reasoning (9999=....) that causes my initial solution (e.g. before my microwave answer) to fail. Can you be more specific why it fails ?
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rmsgrey
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Re: 90>100?
« Reply #15 on: Dec 4th, 2003, 8:34am » |
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Since 9999=3 (mod 7), for several x (eg x=6) x>9999 (mod 7)
Ah! I see what you were getting at - because the mod 7 device thinks everything [in]{0,1,2,3,4,5,6} it definitely thinks anything you give it is smaller than 9999. The problem is that your device doesn't have a concept of "9999" as anything other than "3".
There's also a more subtle problem in that finite fields generally do not have a natural order relation defined upon them. In fact, there is no order relation on the integers mod p which gives an ordered field. Consider the "top" element of the field, t and two elements a[ne]b with a>b. Then a+(t-b)>b+(t-b) but b+(t-b)=t so a+(t-b)>t but t is top. Contradiction.
It being impossible to come up with a proper order on the field of integers mod 7, the interpretation of the inequalities in the problem statement becomes problematic.
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| « Last Edit: Dec 4th, 2003, 4:25pm by Icarus » |
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