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ABCD's Of Greatest Divisor (Posted on 2009-09-24) Difficulty: 3 of 5
The greatest common divisor of five positive integers ABCD, 1920CD41, 496BC3, 872AB76 and 10A25D8 is ≥ 2, where each of A, B, C and D represents a different base 10 digit from 0 to 9.

Determine all possible quadruplet(s) (A, B, C, D) that satisfy the given conditions.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
computer solution | Comment 1 of 3
using mathematica:

For[a=1,a„T9,a++,
For[b=0,b
„T9,b++,
If[b!=
a,
For[c=0,c
„T9,c++,
If[c!=
a && c!=b,
For[d=0,d
„T9,d++,
If[d!
a && d!b && d!c,
v1=FromDigits[{a,b,c,d}];
v2=FromDigits[{1,9,2,0,c,d,4,1}];
v3=FromDigits[{4,9,6,b,c,3}];
v4=FromDigits[{8,7,2,a,b,7,6}];
If[GCD[v1,v2,v3,v4]>=
2,
Print[v1," ",v2," ",v3," ",v4," ",GCD[v1,v2,v3,v4]];
];];];];];];];];

gives the solution of a=8 b=5 c=2 d=4 with GCD of 2131


  Posted by Daniel on 2009-09-24 11:33:15
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