perplexus.info :: Numbers : Pandigital Processed Primes

Pandigital Processed Primes (Posted on 2009-06-27)

Three prime numbers, p₁, p₂, p₃, are processed by multiplying the first by 7, the second by 8 and the third by 9, so you have

7p₁, 8p₂ and 9p₃.

and it's found that

7p₁ < 8p₂ < 9p₃.

Taken together, 7p₁, 8p₂ and 9p₃ use all nine digits 1 - 9 exactly once each, with no zeros.

Also, the product of 7p₁, 8p₂ and 9p₃ is a 9-digit number containing each digit 1 - 9.

What are the three processed primes, 7p₁, 8p₂ and 9p₃?

Submitted by Charlie

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Solution: (Hide)

70 :for A=1 to 12251 80 :Na=prm(A)*7:Nas=cutspc(str(Na)) 90 :if len(Nas)<4 then 100 :for B=1 to 12251 110 :Nb=prm(B)*8:Nbs=cutspc(str(Nb)) 120 :if len(Nbs)>=len(Nas) and 9-len(Nas)-len(Nbs)>=len(Nbs) then 130 :for C=1 to 12251 140 :Nc=prm(C)*9:Ncs=cutspc(str(Nc)) 160 :T$=cutspc(str(Na))+cutspc(str(Nb))+cutspc(str(Nc)) 170 :if len(T$)=9 then 180 :Good=1 190 :for I=1 to 9 200 :if instr(T$,cutspc(str(I)))=0 then Good=0:endif 210 :next 211 :Prods=cutspc(str(Na*Nb*Nc)) 212 :for I=1 to 9 213 :if instr(Prods,cutspc(str(I)))=0 then Good=0:endif 214 :next 220 :if Good then print Na;"= 7 *";Na//7;Nb;"= 8 *";Nb//8;Nc;"= 9 *"; Nc//9,Na*Nb*Nc:endif 230 :endif 250 :next 260 :endif 270 :next 280 :endif 290 :next 300 : finds 413 = 7 * 59 568 = 8 * 71 927 = 9 * 103 where the product is 217459368.

So the numbers are 413, 568 and 927.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Thoughts	K Sengupta	2023-04-24 01:37:10
solution	Daniel	2009-06-28 09:39:37

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