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 Primes in arithmetic progression (Posted on 2008-09-30)
Ten primes, each less than 3000, form an arithmetic progression. Find them.

 See The Solution Submitted by pcbouhid No Rating

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 computer solution | Comment 1 of 9
`   10   repeat   20     N=nxtprm(N):Ct=Ct+1   30   until N>3000   40   NoPrimes=Ct-1   50   for Sub1=1 to NoPrimes-9   60   for Sub2=NoPrimes to Sub1+9 step -1   70     if (prm(Sub2)-prm(Sub1))@9=0 then   80      :Good=1   90      :Increm=int((prm(Sub2)-prm(Sub1))/9)  100      :for I=1 to 8  110        :if prmdiv(prm(Sub1)+I*Increm)<prm(Sub1)+I*Increm then Good=0:endif  120      :next  130      :if Good then  140        :for I=0 to 9  150          :print prm(Sub1)+I*Increm;  151          :P=prm(Sub1)+I*Increm  152          :for J=1 to NoPrimes  153           :if prm(J)=P then print "(";cutspc(str(J));")":endif  154          :next  160        :next  165        :print:print Increm  170      :endif  180   next  190   next`

Finds a sequence starting at 199 in increments of 210:

` 199 (46) 409 (80) 619 (114) 829 (145) 1039 (175) 1249 (204) 1459 (232) 1669 (263) 1879 (289) 2089 (316)`

The program checks all pairs of primes in the given range and sees if they differ by a multiple of 9.  If so, each intervening 1/9 of the way over is checked for primality. If all of them are prime, the result prints.

 Posted by Charlie on 2008-09-30 13:16:46

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