Here's a fairly deterministic proof that doesn't involve much guess-and-checking: Let p be the prices after lunch, let d be the difference in the prices, and l

Here's a fairly deterministic proof that doesn't involve much guess-and-checking: Let p be the prices after lunch, let d be the difference in the prices, and let a, b, c be the number of chickens each farmer sold before and after lunch. Then 3500 = ad + 10p = bd + 16p = cd + 26p. This gives us the following equation: 3500 = d * (a + b - c). Thus d and (a + b - c) must divide 3500. Now, note that (a - b)d = 6p. Thus 3 divides (a - b)d. But d is a divisor of 3500, so 3 cannot divide d, thus 3 must divide a - b. Since 10 >= a > b > c >= 0, this means a = b + 3 or a = b + 6. Similarly 4 must divide a - c, so a = c + 4 or a = c + 8. Finally, (b - c)d = 10p = (5/3)* 6p = (5/3) * (a - b)d. Dividing by d we get b - c = (5/3) * (a - b). Since a - b is 3 or 6, that means b - c is 5 or 10. The only set of possibilities that satisfies (a = c + 4 or c + 8, a = b + 3 or b + 6, b = c + 5 or c + 10, a <= 10) is (a, b, c) = (a, a - 3, a - 8). Since a <= 10, c must be 0, 1, or 2. If c = 0, then we get 3500 = cd + 26p = 26p, which is impossible since 3500 is not divisible by 13. If c = 2, then 3500 = d * (a + b - c) = d * 15, which is impossible since 3500 is not divisible by 3. Thus c = 1, b = 6, a = 9 is the only possibility.