MOTION MOUNTAIN

358 challenge hints and solutions
Challenge 275, page 274: The child is minus 0.75 years old, or minus 9 monthsold;thefatheris
thusvery nearthemother.
Challenge 276, page 274: Thisisnotaneasyquestion.Thefirstnon-trivialnumbersare7,23,47,
59,167and179.SeeRobert Matthews, Maximallyperiodicreciprocals,BulletinoftheInstitute
of Mathematics and its Applications28, pp. 147–148, 1992. Matthewsshowsthat a number
n for which1/n generates the maximum ofn−1decimal digits in the decimal expansion is a
special sortof prime number that can be deduced from the so-called SophieGermain primesS;
onemusthaven=2S+1,wherebothSand2S+1mustbeprimeandwhereSmod20mustbe
3,9,or11.
Thus the first numbersnare 7, 23, 47, 59, 167 and 179,corresponding to values forSof 3, 11,
23,29,83and89.In1992,thelargest knownSthatmeetsthecriteria was
S=(39051⋅2 6002 )−1, (118)
a 1812-digit long SophieGermain prime number thatis3mod 20. It wasdiscovered by Wilfred
Keller. This Sophie Germain prime leads to a primen with a decimal expansion that is around
10 1812 digitslongbeforeitstartsrepeatingitself.Read yourfavouritebookonnumbertheory to
findout more. Interestingly,the solutionto thischallenge isalsoconnected to thatof challenge
267.Canyoufindoutmore?
Challenge 277, page 274: Klein did not belong to either group. As a result, some of his nastier
studentsconcluded thathewasnotamathematicianatall.
Challenge 278, page 274: Abarbercannotbelongtoeithergroup;thedefinitionofthebarberis
thuscontradictoryandhastoberejected.
Challenge 279, page 275: See the members.shaw.ca/hdhcubes/cube_basics.htm web page for
moreinformationonmagiccubes.
Challenge 280, page 275: Such an expression isderived with theintermediate result(1−2 2 ) −1 .
Thehandlingofdivergentseriesseemsabsurd,butmathematiciansknowhowtogivetheexpression
a defined content. (See Godfrey H. Hardy, Divergent Series, Oxford University Press,
1949.) Physicists often use similar expressions without thinking about them, in quantum field
theory.
Challenge 281, page 275: Trytofindanotheroneandthentoprovetheuniquenessoftheknown
one.
Challenge 282, page 276: Theresult isrelated to Riemann’szeta function.Foranintroduction,
seeen.wikipedia.org/wiki/Prime_number.
Challenge 284, page 287: ‘AllCretans lie’ isfalse, sincetheopposite,namely ‘someCretans say
thetruth’ istrueinthecasegiven.Thetrapisthattheoppositeoftheoriginalsentenceisusually,
butfalsely, assumedtobe ‘allCretans saythetruth’.
Challenge 285, page 287: The statement cannot be false, due to the first half and the ‘or’ construction.Sinceitistrue,thesecondhalfmustbetrueandyouareanangel.
Challenge 286, page 287: The terms ‘circular’ and ‘self-referential’ describe two different concepts.
Challenge 288, page 288: Extraterrestrials cannot be at the origin of crop circles because, like
FatherChristmasorghosts,theydonotexistonEarth.
Challenge 290, page 288: Thiscanbedebated;inanycaseitisdefinitelyknownthatbothstatementsarelies,asshownlateron.
Vol. V, page 90 .
Challenge 291, page 289: If this false statement were true, swimmers or divers would also die,
astheirskincannotbreatheeither.
Motion Mountain – The Adventure of Physics copyright © Christoph Schiller June 1990–November 2015 free pdf file available at www.motionmountain.net