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Topic: 51^51 (Read 805 times) |
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SWF
Uberpuzzler
Posts: 879
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Re: 51^51
« Reply #1 on: Feb 21st, 2003, 5:19pm » |
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This problem has been sitting around for a while with no comments at all. As the author of another fine problem which has received no comment after 2 months, I decided to take a shot at this one before it also falls into the abyss of page 2.
Working in modulo 11 arthmetic shows that this can not be expressed as the sum of a square and a fifth power. Straight calculation shows that 5110=1 mod 11. So 5150=(5010)5 is also 1 mod 11, and 5151=7 mod 11.
Trying the eleven possiblities shows that every square is either 0, 1, 3, 4, 5, or 9 mod 11. In the same way, it is found that every fifth power is either 0, 1, or 10 mod 11. Thus it is found that there no way to add a square to a fifth power to get a number equal to 7 mod 11, so 5151 cannot be expressed in this way.
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Posts: 4863
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Re: 51^51
« Reply #2 on: Feb 22nd, 2003, 12:02pm » |
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I've been playing with this for some time without making any progress. Out of curiousity, SWF, how did you come up with this approach?
Concerning your orphaned puzzle - sorry, but that sort of thing doesn't have much appeal to me - but for those with different tastes, here is a link.
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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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SWF
Uberpuzzler
Posts: 879
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Re: 51^51
« Reply #3 on: Feb 23rd, 2003, 1:05pm » |
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I guessed that it was not possible, probably because if it was, it looked to be difficult. Without doing an exhaustive search, looking at modular arithmetic seemed like the best way to show it impossible. I figured a proof of impossibility would be much as written above, except I didn't know what number to use as the modulus. I just tried looking at possibilities for squares and fifth powers with different moduli until finding that modulo 11 had some impossible to reach values for the sum. Also, if it was possible, modular arithmetic would have given clues as to characteristics of the answer, which would have helped reduce the cases to check if it ever came to an exhaustive search.
As for my orphaned problem, I consider it being neglected as tribute to its difficulty. Although one might expect somebody from this forum would be able to solve a 13 piece puzzle.
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Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
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Posts: 4863
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Re: 51^51
« Reply #4 on: Feb 24th, 2003, 4:12pm » |
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In hindsight, 11 would be the best modulus to start looking with, since Fermat's little theorem provides n10=1 mod 11, if (n,11)=1, and both 2 and 5 divide 10, this limits the number of possible squares and 5th powers modulo 11. I'm guessing that you could also use mod 31 or mod 41. It's amazing how easy it is to see this if you already know the answer!
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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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David Ryan
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Would anyone be willing to expain to me what modulus is? I have never heard of it, and don't foresee it surfacing reletively soon in math class.
If it is not practical here, that is fine, please say so.
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wowbagger
Uberpuzzler
 
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Re: 51^51
« Reply #6 on: Mar 7th, 2003, 8:20am » |
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David,
when using "modular arithmetic", you basically throw away integral multiples of some number m, the modulus.
Example: 7 = 1 (mod 6)
Since 7 is greater than 6, you subtract 6 (once in this case) and end up with 1.
For more info, let me refer you to a good web resource on mathematical terms: World of Mathematics with definitions of modulus and the related concept of congruence.
SWF,
please excuse my not posting to your puzzle. I cut out the pieces and really tried to solve it for a while (back in December and January), but didn't get very far. So I thought it's not worth a post.
Your creation did not go unnoticed, I can assure you. 
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| « Last Edit: Mar 7th, 2003, 8:30am by wowbagger » |
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