exam1 - M ath 470/501 Spring 2011 Exam 1 M aple allowed...

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Unformatted text preview: M ath 470/501 Spring 2011 Exam 1 M aple allowed Name: U IN: #1. The title of this 1966 musical has been encrypted by a shift cipher: For $200, Watson, what is its name? > restart: with(LinearAlgebra): with(numtheory): with(plots): numb:=table([""=0,"a"=1,"b"=2,"c"=3,"d"=4,"e"=5,"f"=6,"g"=7, "h"=8,"i"=9,"j"=10,"k"=11,"l"=12,"m"=13,"n"=14,"o"=15,"p"=16, "q"=17,"r"=18,"s"=19,"t"=20,"u"=21,"v"=22,"w"=23,"x"=24,"y"=25, "z"=26]): alph:=table([0="",1="a",2="b",3="c",4="d",5="e",6="f",7="g",8= "h",9="i",10="j",11="k",12="l",13="m",14="n",15="o",16="p",17= "q",18="r",19="s",20="t",21="u",22="v",23="w",24="x",25="y",26= "z"]): shift:=proc(txt,n) local i,z; # remove the # on only the following line to trace the internal outputs # option trace; # Create a NULL string on which to add letters. z:=NULL; # While i is less than or equal to the number of letters in the string, do for i from 1 while i<=length(txt) do # Grab the substring (letter) in the i-th place of txt. substring(txt,i); # Use the numb table to convert the letter to a n umber between 1 and 26. numb[%]; # Arithmetic mod 26 uses numbers between 0 and 25, not 1 and 26 %-1; # Now cyclically shift the letter mod 26. %+n mod 26; # Take the numbers between 0 and 25 and again make them between 1 and 26, as people count letters. %+1; # Use the alph table to convert the shifted number back to a letter. alph[%]; # Add the new letter on the right side of the shifted string. z:=cat(z,%); end do; return(z); # To check that syntax is correct, change end: to end; Blue output means "syntax is OK." end: allshifts:= proc(txt) local i; for i from 0 while i<26 do shift(txt,26+i); print(%); end do; end: mu s i c a l:=" te d lmtc o te d lawlypt eddf apcxly"; allshifts(musical); "tedlmtcotedlawlypteddfapcxly" "ufemnudpufembxmzqufeegbqdymz" "vgfnoveqvgfncynarvgffhcrezna" "whgopwfrwhgodzobswhggidsfaob" "xihpqxgsxihpeapctxihhjetgbpc" "yjiqryhtyjiqfbqduyjiikfuhcqd" "zkjrsziuzkjrgcrevzkjjlgvidre" "alkstajvalkshdsfwalkkmhwjesf" "bmltubkwbmltietgxbmllnixkftg" "cnmuvclxcnmujfuhycnmmojylguh" "donvwdmydonvkgvizdonnpkzmhvi" "epowxenzepowlhwjaepooqlaniwj" "fqpxyfoafqpxmixkbfqpprmbojxk" "grqyzgpbgrqynjylcgrqqsncpkyl" "hsrzahqchsrzokzmdhsrrtodqlzm" "itsabirditsaplaneitssuperman" "jutbcjsejutbqmbofjuttvqfsnbo" "kvucdktfkvucrncpgkvuuwrgtocp" "lwvdeluglwvdsodqhlwvvxshupdq" "mxwefmvhmxwetperimxwwytivqer" "nyxfgnwinyxfuqfsjnyxxzujwrfs" " ozyghoxjozygvrgtkozyyavkxsgt" "pazhipykpazhwshulpazzbwlythu" "qbaijqzlqbaixtivmqbaacxmzuiv" "rcbjkramrcbjyujwnrcbbdynavjw" "sdcklsbnsdckzvkxosdccezobwkx" (1.1) (1.1) #2. After the pair 2, 3, which are separated by 1, all further primes must be s eparated by at least two. 3 and 1223 are primes. What fraction of the primes p b etween 3 and 1223 are twin primes, meaning that p is prime and p+2 is also p rime? Write a short loop in Maple to determine the answer. Execute your loop. P rint your output. > p:=0: t:=0: for i from 3 to 1223 do if isprime(i) then p:=p+1; if isprime(i+2) then t:=t+1; #print(i,i+2); fi; fi; od; t/p; evalf(%); 41 199 0.2060301508 (2.1) ...
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This note was uploaded on 02/14/2012 for the course MATH 470 taught by Professor Staff during the Spring '08 term at Texas A&M.

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