Re: [math-fun] "Ponder This"
Allan Wechsler <acwacw@gmail.com> wrote:
Dan Asimov <dasimov@earthlink.net> wrote:
ELEVEN + TWO = TWELVE + ONE
Dan, this is just an anagram, right? Unless I'm missing something truly devilish, it's arithmetically impossible.
I've confirmed that there is no solution. Since there are only seven distinct letters, I also tried base 7. Then all bases through 40. Still no solution. (Larger bases are very slow to test.) Would one expect there to be a (possibly non-unique) solution in some base, eventually, just by chance? If so, I propose the opposite challenge -- finding an "anti-alphametic," an arrangement that has no solution in any base. Getting back to my original puzzle under this subject line, I'm about halfway through searching the six-four case, ______=____ with interspersed * + - . / So far I've found about 430,000 arrangements with non-unique solutions and 220 with unique solutions. Of those 220, just 2 have negative values: __*_._-_._=_/__-_ and _*_._/_-_/_=_-__._ Of those 220, 13 contain all of * + - . / and 4 contain all of those once and only once. Those 4 are closely related so I'll only give one of them: __*_.___=_/_-_+_ Again, you're to replace the underscores with each of the ten digits once and only once, and there is a unique solution. I'll re-ask my question: Given that I'm never simplifying fractions, and that I cross-multiply them at the end to check for equality, what are the largest numbers that could appear? Am I at risk for integer overflow? For instance _*_.____=__-_/_ has the unique solution 4*2.8750=13-9/6, which yeilds the fractions 115000/10000 and 69/6. Cross-multiplying those fractions to check if they're equal gives 115000*6 = 10000*69 = 690000.
On 4/21/2013 9:50 AM, Keith F. Lynch wrote:
Allan Wechsler <acwacw@gmail.com> wrote:
Dan Asimov <dasimov@earthlink.net> wrote:
ELEVEN + TWO = TWELVE + ONE
Dan, this is just an anagram, right? Unless I'm missing something truly devilish, it's arithmetically impossible.
I've confirmed that there is no solution. Since there are only seven distinct letters, I also tried base 7. Then all bases through 40. Still no solution. (Larger bases are very slow to test.)
There is no solution in any base. The matching Es in the 1000s place are the trouble: The difference between ELE000 and TWE000 is a nonzero multiple of 10000, but it is also |VEN + TWO - LVE - ONE|, which is less than 2000. -- Fred W. Helenius fredh@ix.netcom.com
Related to this topic are some finds I made when exploring a system that interprets words (or letter strings) as numbers written in a positional notation using base 27. As usual, A=1, B=2, etc., with zero represented by an underscore. Hence CAT = 3*27^2 + 1*27^1 + 20*27^0 = 2234. The finds I refer to are congruences that are true on two levels: (FORTYTWO + FORTYNINE mod SIX <-> ONE and (42+49) mod 6 <-> 1, (EIGHTYFIVE - EIGHTYTWO) mod TWO <-> ONE and (85-82) mod 2 <-> 1, (NINETYNINE - TWELVE) mod FIVE <-> TWO and (99-12) mod 5 <-> 2. Of course, equations would be far preferable to congruences, if such could be found. Anyone like to try? Lee At 05:14 PM 4/21/2013, you wrote:
On 4/21/2013 9:50 AM, Keith F. Lynch wrote:
Allan Wechsler <acwacw@gmail.com> wrote:
Dan Asimov <dasimov@earthlink.net> wrote:
ELEVEN + TWO = TWELVE + ONE
Dan, this is just an anagram, right? Unless I'm missing something truly devilish, it's arithmetically impossible.
I've confirmed that there is no solution. Since there are only seven distinct letters, I also tried base 7. Then all bases through 40. Still no solution. (Larger bases are very slow to test.)
There is no solution in any base. The matching Es in the 1000s place are the trouble: The difference between ELE000 and TWE000 is a nonzero multiple of 10000, but it is also |VEN + TWO - LVE - ONE|, which is less than 2000.
-- Fred W. Helenius fredh@ix.netcom.com
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Hi Lee, its always a pleasure to read you! Here is the French equivalent of ELEVEN +... ... which has far less elegance: http://www.cetteadressecomportecinquantesignes.com/Eleven.htm This was found years ago by a guy who shared my body for a long time... Melvin O. is quoted at the bottom of the page... Souvenirs, souvenirs... Best, É. Propulsé d'un aPhone Le 21 avr. 2013 à 17:48, "Lee Sallows" <Lee.Sal@inter.nl.net> a écrit :
Related to this topic are some finds I made when exploring a system that interprets words (or letter strings) as numbers written in a positional notation using base 27. As usual, A=1, B=2, etc., with zero represented by an underscore. Hence CAT = 3*27^2 + 1*27^1 + 20*27^0 = 2234.
The finds I refer to are congruences that are true on two levels:
(FORTYTWO + FORTYNINE mod SIX <-> ONE and (42+49) mod 6 <-> 1, (EIGHTYFIVE - EIGHTYTWO) mod TWO <-> ONE and (85-82) mod 2 <-> 1, (NINETYNINE - TWELVE) mod FIVE <-> TWO and (99-12) mod 5 <-> 2.
Of course, equations would be far preferable to congruences, if such could be found. Anyone like to try?
Lee
At 05:14 PM 4/21/2013, you wrote:
On 4/21/2013 9:50 AM, Keith F. Lynch wrote:
Allan Wechsler <acwacw@gmail.com> wrote:
Dan Asimov <dasimov@earthlink.net> wrote:
ELEVEN + TWO = TWELVE + ONE
Dan, this is just an anagram, right? Unless I'm missing something truly devilish, it's arithmetically impossible.
I've confirmed that there is no solution. Since there are only seven distinct letters, I also tried base 7. Then all bases through 40. Still no solution. (Larger bases are very slow to test.)
There is no solution in any base. The matching Es in the 1000s place are the trouble: The difference between ELE000 and TWE000 is a nonzero multiple of 10000, but it is also |VEN + TWO - LVE - ONE|, which is less than 2000.
-- Fred W. Helenius fredh@ix.netcom.com
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On Sun, Apr 21, 2013 at 9:50 AM, Keith F. Lynch <kfl@keithlynch.net> wrote:
Would one expect there to be a (possibly non-unique) solution in some base, eventually, just by chance? If so, I propose the opposite challenge -- finding an "anti-alphametic," an arrangement that has no solution in any base.
The simplest one is A + A = AA; in base B, A < B, so A + A < 2B, so the carry is 1, so A = 1. 1+1 results in a carry only in base 2, which doesn't work, so we're done. Andy
participants (5)
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Andy Latto -
Eric Angelini -
Fred W. Helenius -
Keith F. Lynch -
Lee Sallows