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My solution is below. Federico Kereki provides a good solution here.
Let S be the number of Smiths, A be the number of Andrings and C be the number of Cliffords.
From the problem: 0<S<=A<=C and SA + AC + SC = 142.
Combining the two equations gives an upper bound for S: 3S^2 <= 142. S<7.
SA + AC + SC = 142 can be rewritten as (A+S)*(C+S) = 142 + S^2.
Case S=1:
(A+1)*(C+1) = 143
A=10 and C=12
Case S=2:
(A+2)*(C+2) = 146
No factors give A,C positive
Case S=3:
(A+3)*(C+3) = 151
No factors give A,C positive
Case S=4:
(A+4)*(C+4) = 158
No factors give A,C positive
Case S=5:
(A+5)*(C+5) = 167
No factors give A,C positive
Case S=6:
(A+6)*(C+6) = 178
No factors give A,C positive
The only solution in positive integers is S=1, A=10, C=12.
There were 12 Cliffords, 10 Andrings and only one Smith. |
