MOTION MOUNTAIN

challenge hints and solutions 357
Vol. V, page 308
Vol. VI, page 142
Challenge 256, page 259: Physicists claim that the properties of objects, of space-time and of
interactions form the smallest list possible. However, this list is longer than the one found by
linguists!Thereasonisthatphysicistshavefoundprimitivesthatdonotappearineveryday life.
Inasense,theaimof physicistsislimited by listof unexplained questionsof nature,given later
on.
Challenge 257, page 261: Neither has a defined content, clearly stated limits or a domain of
application.
Challenge 258, page 261: Impossible! That would not be a concept, as it has no content. The
solutiontotheissuemustbeandwillbedifferent.
Challenge 259, page 263: To neither. This paradox shows that such a ‘set of all sets’ does not
exist.
Challenge 260, page 263: Themostfamousistheclassofallsetsthatdonotcontainthemselves.
Thisisnotaset,butaclass.
Challenge 261, page 264: Dividing cakes is difficult. A simple method that solves many – but
notall–problemsamongNpersonsP1...PNisthefollowing:
— P1cutsthecake intoNpieces.
— P2toPNchooseapiece.
— P1keeps thelastpart.
— P2...PNassembletheirpartsbackintoone.
— ThenP2...PNrepeatthealgorithm foronepersonless.
The problem is much more complex if the reassembly is not allowed. A just method (in finite
manysteps)for3people,usingninesteps,waspublishedin1944bySteinhaus,andafullysatisfactory
method in the 1960s by John Conway. A fully satisfactory method for four persons was
foundonlyin1995;ithas20steps.
Challenge 262, page 264:(x,y):={x,{x,y}}.
Challenge 263, page 265: Hint:showthatanycountablelistofrealsmissesatleastonenumber.
Thiswas proven forthefirst timeby Cantor.Hisway wasto write thelistindecimal expansion
andthenfindanumberthatissurelynotinthelist.Secondhint:hisworld-famoustrickiscalled
thediagonalargument.
Challenge 264, page 265: Hint:allreals arelimitsofseriesofrationals.
Challenge 266, page 267: Yes, but only provided division by zero is not allowed, and numbers
arerestricted totherationalsandreals.
Challenge 267, page 267: There are infinitely many of these so-called parasitic numbers. The
smallest is already large: 1016949152542372881355932203389830508474576271186440677966.
If the number 6 is changed in the puzzle, one finds that the smallest solution for 1 is
1, for 4 is 102564, for 5 is 142857, for 8 is 1012658227848, for 2 is 105263157894736842,
for 7 is 1014492753623188405797, for 3 is 1034482758620689655172413793, and for 9 is
10112359550561797752808988764044943820224719. The smallest solution for 6 is the largest
ofthislist.
Challenge 268, page 267: Onewaywasgivenabove:0:= , 1:={⌀} , 2:={{⌀}} etc.
Challenge 269, page 271: Subtractioniseasy.Additionisnotcommutativeonly forcases when
infinitenumbersareinvolved:ω+2=2+ω.
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Challenge 270, page 271: Examplesare1−εor1−4ε 2 −3ε 3 .
Challenge 271, page 272: Theansweris57;thecitedreference gives thedetails.
Challenge 272, page 274:2 222 and4 444 .
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Motion Mountain – The Adventure of Physics copyright © Christoph Schiller June 1990–November 2015 free pdf file available at www.motionmountain.net