Took me about 20 minutes to solve using a bruteforce solution scaling linearly over the total revenue (in cents, so in this case 10500 iterations). Writing dow

Took me about 20 minutes to solve using a bruteforce solution scaling linearly over the total revenue (in cents, so in this case 10500 iterations). Writing down equations: x*a + (10 - x)*b = 3500 y*a + (16 - y)*b = 3500 z*a + (26 - z)*b = 3500 x*a + (10 - x)*b + y*a + (16 - y)*b + z*a + (26 - z)*b = 3 * 3500 (x + y + z)*a + (10 - x + 16 - y + 26 - z)*b = 10500 (x + y + z)*a + (52 - (x + y + z))*b = 10500 p = x + y + z p*a + (52 - p)*b = 10500 a = (10500 - (52 - p)*b) / p p is the total number of chickens sold before the break. Obviously this can't be negative nor more than 52. At least 1 chicken must be sold both before and after the break, because 52 is not a factor of 10500. So 0 < p < 52. We know that b < a so b < (10500 - (52 - p) * b) / p. b < (10500 - (52 - p)*b) / p b*p < 10500 - (52 - p)*b b*p + (52 - p)*b < 10500 b*52 < 10500 b < 202 This is easy to bruteforce: for p in range(1, 52): for b in range(1, 202): if (10500 - (52 - p)*b) % p == 0: a = (10500 - (52 - p)*b) // p if ((3500 - 10*b) % (a - b) == 0 and (3500 - 16*b) % (a - b) == 0 and (3500 - 26*b) % (a - b) == 0): x = (3500 - 10*b) // (a - b) y = (3500 - 16*b) // (a - b) z = (3500 - 26*b) // (a - b) if 0 <= x <= 10 and 0 <= y <= 16 and 0 <= z <= 26: print("Before the break one chicken cost ${:.2f} and the farmers sold ({}, {}, {}).".format(a / 100, x, y, z)) print("After the break one chicken cost ${:.2f} and the farmers sold ({}, {}, {}).".format(b / 100, 10-x, 16-y, 26-z)) Spoiler: Before the break one chicken cost $3.75 and the farmers sold (9, 6, 1). After the break one chicken cost $1.25 and the farmers sold (1, 10, 25).