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Solution to Enigma No. 1599

9 June 2010

Answer: 1, 1, 3 (these may be listed in any order)

The winner is J D S Bullock of Gainsborough, Lincolnshire, UK.

Worked solution

The digits used can only come from 1, 3, 7 and 9. Any number that has three identical digits is divisible by 3.

If the three digits are all different:

1, 3, 7: 371 = 7×53 713 = 23×31 731 = 17×43

1, 3, 9: 319 = 11×29 391 = 17×23 913 = 11×83 931 = 7x7x19

1, 7, 9: 791 = 7×113 917 = 7×131

3, 7, 9: 793 = 13×61 973 = 7×139

If two digits are the same the following can be eliminated because all their numbers are divisible by 3: 1, 1, 7; 3, 3, 9; 7, 7, 1; 9, 9, 3.

1, 1, 3: 113, 131, 311 are all prime.

1, 1, 9: 119 = 7×17     

3, 3, 1: 133 = 7×19     

3, 3, 7: 337, 373, 733 are all prime.

7, 7, 3: 377 = 13×29 737 = 11×67

7, 7, 9: 779 = 19×41     

9, 9, 1: 199, 919, 991 are all prime.

9, 9, 7: 799 = 17×47 979 = 11×89

So the only examples are 1, 1, 3; 3, 3, 7; 1, 9, 9.

The only one of these that has a digit in common with each of the other two is 1, 1, 3.

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