osman@roxie.DEC (Eric, Digital, Maynard, 617 493-6664) (02/11/86)
Start now ! /Eric
matt@oddjob.UUCP (Matt Crawford) (02/13/86)
The subject line is long enough already! Unless I forget, the
following will be a rot13 solution:
Snpg 1: nyy ahzoref va gur frg ner rdhny. Vs gurl jrera'g,
gura nal cnve bs hardhny ahzoref {n, o} va gur frg pbhyq or
ercynprq ol { (n+o)/2, (n+o)/2 } tvivat n ynetre cebqhpg jvgu
gur fnzr fhz. (Jbex bhg gur znkvzvmngvba ceboyrz sbe n*o
fhowrpg gb n+o=p. Lbh trg n=o.)
Gurersber, gur frg bs ahzoref (npghnyyl n "ont") jvyy or a
pbcvrf bs 100/a, sbe fbzr a. Gerngvat a nf n pbagvabhf inevnoyr
naq znkvzvmvat (100/a)^a tvirf a=100/r ~= 36.79, fb gur nafjre
jvyy or a=36 be a=37. Gelvat gurz obgu fubjf gung a=37 vf gur
nafjre naq gur cebqhpg vf nobhg 9.400r15.
The pure mathemetician will criticize some of the glib
shortcuts, but this is how a physicist solves it. The answer is
right and it took only 4 minutes, so think before you flame!
_____________________________________________________
Matt University crawford@anl-mcs.arpa
Crawford of Chicago ihnp4!oddjob!mattldenenbe@bbncc5.UUCP (Larry Denenberg) (02/13/86)
It can't be done.
Let S be any set of positive real numbers whose sum is 100. If S has one
element then it is {100} and the set {49,51} has greater product. Otherwise
S has two elements x and y with x < y. Choose any z such that 0 < z < y-x
and neither x+z nor y-z is in S---this is obviously possible if S is countable
(but if not then S has an infinite number of elements less than 1 and its
product is as close to zero as makes no never mind). Replace x and y in S
with x+z and y-z. The sum of S is still 100, and the product has increased
since (x+z)(y-z) = xy + (y-x)z - zz > xy.
OK, how close can we get? Well, the argument above, stripped of formality,
just says that if two numbers in S are different you can improve the product
by bringing them closer together. If S were not a set but a multiset, this
process could stop only when all the elements in S were equal. If we look at
the function x**(100/x) we find (not surprisingly) that x = e gives the
global maximum. So S should have 100/e copies of e. But 100/e is not an
integer; it's between 36 and 37. The best multiset is thus either
36 copies of 100/36 or 37 copies of 100/37; probably the latter, since
100/e is 36.78 or so. To get a good set S, just take 37 copies of 100/37
and perturb them a hair: pick z = 10**(-10000000000), say, and let
S = { 100/37 - z, 100/37 + z, 100/37 - 2z, 100/37 + 2z, ... }
where S has 37 elements and the coefficients of the z's add up to 0. This
gives a good solution, but picking a smaller z will always give a better one.
/Larry Denenberg
larry@{bbncc5.ARPA,harvard.harvard.EDU}greg@harvard.UUCP (Greg) (02/13/86)
In article <1692@bbncc5.UUCP> larry@bbncc5.UUCP (Larry Denenberg) writes: >If S were not a set but a multiset, this >process could stop only when all the elements in S were equal. If we look at >the function x**(100/x) we find (not surprisingly) that x = e gives the >global maximum. So S should have 100/e copies of e. But 100/e is not an >integer; it's between 36 and 37. The best multiset is thus either >36 copies of 100/36 or 37 copies of 100/37; probably the latter, since >100/e is 36.78 or so. Fixing a number n, we are choosing the multiset {100/n, 100/n, ..., 100/n} and taking the product of the members, which is (100/n)^n. If we wish to maximize this, we may simply maximize its logarithm, which is n*(log(100)-log(n)). Well, 36*(log(100)-log(36))=36.779445, and 37*(log(100)-log(37))=36.787334, so the second one wins. Funny that the maximum of x*(log(100)-log(x)) should occur when x = x*(log(100)-log(x)). -- gregregreg
johansen@agrigene.UUCP (02/15/86)
> Start now ! > > /Eric 49 51 ! /Eric (a different one)
greg@unlv.UUCP (Greg Wohletz) (02/17/86)
In article <1057@decwrl.DEC.COM> osman@roxie.DEC (Eric, Digital, Maynard, 617 493-6664) writes: >Start now ! > >/Eric How about 3.030303... 33 times which yields 7.7453*10^15 --Greg seismo!unrvax!unlv!greg