Men, Wives... Sheep
Three men, Arthur, Brian, and Charles, went with their wives to an auction market to buy some sheep.
Their wives' names were Rachel, Stef, and Tracy, not necessarily in that order.
The average price (in dollars) that each person paid for their sheep was the same as the actual number of sheep that they bought. For example, if Arthur bought sheep at dollars each, he spent dollars altogether.
Arthur bought 23 more sheep than Stef, Brian bought 11 more sheep than Rachel, and each man spent 63 dollars more than his wife.
Who is married to whom?
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1 solution
Let us call the sheep bought by Arthur, Brian, Charles, Rachel, Stef and Tracy: A, B, C, R, S and T.
A - 23=S B – 11=R A2=Wife of A2+63 same for B and C
We can rearrange A2=Wife of A2+63 to get: A2-Wife of A2=63. This factorizes to: (A-Wife of A)(A+Wife of A)=63.
Two integers must multiply together to get 63: these could be 1 and 63, 3 and 21 or 7 and 9.
Therefore, x-Wife of x=1 and x+Wife of x=63. Solving these simultaneously we get 32 and 31.
For another x, x-Wife of x=3 and x+Wife of x=21. Solving these simultaneously we get 9 and 12.
For a final x, x-Wife of x=7 and x+Wife of x=9. Solving these simultaneously we get 1 and 8.
A/B/C=32/12/8 R/S/T=31/9/1
Of the possible values for B and R, the only ones with difference 11 are 12 and 1 so B=12 and R=1. The wife of B is 9 and the husband of R is 8.
The possible values with a difference of 23 for A and S are 32 and 9. So A=32 and S=9. The wife of A is 31 and the husband of S is 12.
We can now link people up: B is the husband of S 122-92=63 C is the husband of R 82-12=63 A is the husband of T 322-312=63
So to summarize – Arthur is married to Tracy, Brian is married to Stef and Charles is married to Rachel.