May 2015 - Archimedean Cattle (Solution)
The answer is 150. Call the number of white bulls and cows WB and WC respectively. Call the number of black bulls and cows BB and BC respectively. Call the number of yellow bulls and cows YB and YC respectively.
The clues can be rewritten as the following:
1: 3WB = BB + YB
Now, by clue 2, YC = 2WC - YB. Thus clue 4 can be rewritten as 3BC = WB + WC + (2WC - YB) = WB + 3WC - YB. Also, by clue 5, BC = 2YB - WC - BB. Thus clue 4 can be rewritten as 3(2YB - WC - BB) = 6YB - 3WC - 3BB = WB + 3WC - YB. This simplifies to 7YB = 3BB + WB + 6WC. Now, by clue 1, YB = 3WB - BB. Thus clue 3 can be rewritten as 2BB = WB + (3WB - BB) so 3BB = 4WB. Thus clue 4 can be rewritten as 7(3WB - BB) = 21WB - 7BB = 4WB + WB + 6WC = 5WB + 6WC. Thus 16WB = 7BB + 6WC. However, since 4WB = 3BB, this simplifies to 12BB = 7BB + 6WC so 5BB = 6WC. Let BB = 12n for some integer n. Since 5BB = 6WC, WC = 10n. Also, since 3BB = 4WB, WB = 9n. Now, YB =3WB - BB = 27n - 12n = 15n. Also, BC = 2YB - WC - BB = 30n - 10n - 12n = 8n. Finally, YC = 2WC - YB = 20n - 15n = 5n. To recap, WB = 9n, WC = 10n, BB = 12n, BC = 8n, YB = 15n, and YC = 5n. Now consider rule 6. WB + BC = 9n + 8n = 17n. However, 17n = x2 - 1 = (x+1) (x-1). Since 17 is prime, at least one of x+1 and x-1 must be divisible by 17. The smallest value of n is obtained by having x+1 = 17 and x-1 = n. Now, n = 15 so WC = 10n = 150.
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