Solution to Enigma No. 1552

1 July 2009

Answer: The digits are 1, 7 and 8.

The winner is Mikey Teal of Acaster Malbis, York, UK. There were 243 entries.

Worked answer

The digits cannot include 0 because any multiple of 10 is the product of more than two primes.

Also, since any number that ends in 24, 28, 48, 64 or 68 is a multiple of 2², these combinations of even digits cannot be used, leaving only 2 and 6 among even digits; but since any number that ends in 12, 16, 32, 36, 52, 56, 72, 76, 92 or 96 is a multiple of 2² these digits cannot be used.

So the three digits are either all odd, or two odd with 4 or 8.

All odd:

1/5/7; 3/5/7; and 5/7/9: impossible because numbers that end in 75 are multiples of 5².
1/3/5: impossible because all the numbers are multiples of 3².
1/3/7: 137, 173, 317 are all prime.
1/3/9: 139, 193 are both prime; 931 = 7² x 19.
1/5/9: 195 = 3 x 5 x 13; 915 = 3 x 5 x 61.
1/7/9: 179, 197, 719, 971 are all prime.
3/5/9: 359, 593, 953 are all prime; 539 = 7² x 11; 935 = 5 x 11 x 17.
3/7/9: 379, 397, 739, 937 are all prime.

Two odd with 4 or 8:

1/7/4; 3/5/4; 1/3/8; and 7/9/8: impossible because all the numbers are multiples of 3 and in each case the two that end in the even digit are multiples of both 2 and 3.
5/7/4; 5/7/8: impossible because numbers that end in 75 are multiples of 5².
5/9/4; 1/9/8; 3/7/8: impossible because all the numbers are multiples of 3².
1/3/4: 431 is prime.
1/5/4: 541 is prime; 154 = 2 x 7 x 11.
1/9/4: 149, 419, 491, 941 are all prime.
3/7/4: 347, 743 are both prime; 374 = 2 x 11 x 17.
3/9/4: 349, 439 are both prime.
7/9/4: 479, 947 are both prime.
1/5/8: 518 = 2 x 7 x 37.
1/7/8: 178 = 2 x 89; 187 = 11 x 17; 718 = 2 x 359; 781 = 11 x 71; 817 = 19 x 43; 871 = 13 x 67.
3/5/8: 853 is prime; 385 = 5 x 7 x 11.
3/9/8: 389, 839, 983 are all prime; 938 = 2 x 7 x 67.
5/9/8: 859 is prime; 598 = 2 x 13 x 23.