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?

I used the following mathematica code

dgts:={1,2,3,4,5,6,7,8,9};

prms=Permutations[dgts];

lng=Length[prms];

cnt=0;

For[i=1,i„Tlng,i++,

num=FromDigits[prms[[i]]];

If[Mod[num,9*8*7]ƒú0,

p1p2p3=num/(9*8*7);

facs=FactorInteger[p1p2p3];

lng2=Length[facs];

tot=Sum[facs[[j,2]],{j,1,lng2}];

If[tot==3,

p1=facs[[1,1]];

facs=FactorInteger[p1p2p3/p1];

p2=facs[[1,1]];

p3=p1p2p3/(p1*p2);

ary={{p1,p2,p3},{p1,p3,p2},{p2,p1,p3},{p2,p3,p1},{p3,p1,p2},{p3,p2,p1}};

For[j=1,j„T6,j++,

v1=7*ary[[j,1]];

v2=8*ary[[j,2]];

v3=9*ary[[j,3]];

v1d=IntegerDigits[v1];

v2d=IntegerDigits[v2];

v3d=IntegerDigits[v3];

digun=Union[v1d,v2d,v3d];

If[v1<v2 && v2<v3 && Length[v1d]+Length[v2d]+Length[v3d]ƒúLength[digun] && digun==dgts,

Print[p1," ",p2," ",p3];

Print[7*p1," ",8*p2," ",9*p3];

Print[7*8*9*p1*p2*p3];

];];];];];

and got the answer of p1=59 p2=71 and p3=103 which gives for 7p1,8p2,9p3

413, 568, 927

and for 7*8*9*p1*p2*p3

217459368

Posted by Daniel on 2009-06-28 09:39:37