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si a,b,c sont les 3 chiffres, on a a(10-c)=b(c-1)
en faisant varier c, je trouve 4 solutions:
   1,2,4   1,2,8   2,3,7   3,4,6
3 solutions supplémentaires, si on admet 2 chiffres identiques:
   2,4,4   4,4,8   4,5,5

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(a+b)*c = 10a+b
ac+bc = 10a+b
bc-b = 10a-ac
b(c-1) = a(10-c); b est à 10-c ce que a est à c-1
c = 2, c-1 = 1, 10-c = 8, a = 1, b = 8; (1+8)*2 = 18
c = 3, c-1 = 2, 10-c = 7, a = 2, b = 7; (2+7)*3 = 27
c = 4, c-1 = 3, 10-c = 6, a = 1, b = 2; (1+2)*4 = 12
                          a = 2, b = 4; (2+4)*4 = 24
                          a = 3, b = 6; (3+6)*4 = 36
                          a = 4; b = 8; (4+8)*4 = 48
c = 5, c-1 = 4, 10-c = 5, a = 4; b = 5; (4+5)*5 = 45
c = 6, c-1 = 5, 10-c = 4, a = 5, b = 4; (5+4)*6 = 54
c = 7, c-1 = 6, 10-c = 3, a = 2, b = 1; (2+1)*7 = 21
                          a = 4, b = 2; (4+2)*7 = 42
                          a = 6, b = 3; (6+3)*7 = 63
                          a = 8, b = 4; (8+4)*7 = 84
c = 8, c-1 = 7, 10-c = 2, a = 7, b = 2; (7+2)*8 = 72
c = 9, c-1 = 8, 10-c = 1, a = 8, b = 1; (8+1)*9 = 81