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 Archimedes' Cattle (Posted on 2004-02-17)
The sun god had a herd of cattle consisting of bulls and cows, one part of which was white, a second black, a third spotted, and a fourth brown.

Among the bulls, the number of white ones was one half plus one third the number of the black greater than the brown; the number of the black, one quarter plus one fifth the number of the spotted greater than the brown; the number of the spotted, one sixth and one seventh the number of the white greater than the brown.

Among the cows, the number of white ones was one third plus one quarter of the total black cattle; the number of the black, one quarter plus one fifth the total of the spotted cattle; the number of spotted, one fifth plus one sixth the total of the brown cattle; the number of the brown, one sixth plus one seventh the total of the white cattle.

What was the composition of the herd?

 Submitted by DJ Rating: 3.5000 (8 votes) Solution: (Hide) ``` 10,366,482 white bulls 7,460,514 black bulls 7,358,060 spotted bulls 4,149,387 brown bulls 7,206,360 white cows 4,893,246 black cows 3,515,820 spotted cows 5,439,213 brown cows ``` The first task is to convert all of that into equations. Using W, X, Y, Z, w, x, y, z for white bulls, black bulls, spotted bulls, brown bulls, white cows, black cows, spotted cows, and brown cows, respectively, we get these equations: ```W = (5/6)X + Z X = (9/20)Y + Z Y = (13/42)W + Z w = (7/12)(X+x) x = (9/20)(Y+y) y = (11/30)(Z+z) z = (13/42)(W+w) ``` Combining the first three equations, we get: ```W = (5/6)((9/20)Y+Z)+Z = (5/6)((9/20)((13/42)W+Z)+Z)+Z = (13/112)W + (3/8)Z + (5/6)Z + Z = (13/112)W + (53/24)Z so (99/14)W = (53/3)Z and 297W = 742Z. ``` This is reduced to its smallest values, so W is divisible by 742 and Z is divisible by 297. By equation #3, W is also divisible by 42. 2226 (3x742) is the smallest number divisible by both 742 and 42, so W is divisible by 2226. Let's try some W's: ``` W = 2226 4452 6677 8903 ... X = 1602 3204 4806 6408 ... Y = 1580 3160 4740 6320 ... Z = 891 1782 2673 3564 ... ``` The second column is twice the first; the third is three times the first, etc. For each of these columns, we can plug in the values of W, X, Y, and Z into equations #4 through #7. So we have four equations with four unknowns, and we should be able to solve for w, x, y, and z. Choosing the first column, we get these four equations: ```w = (7/12)(1602+x) x = (9/20)(1580+y) y = (11/30)(891+z) z = (13/42)(2226+w) ``` Solving these four equations for z, we get: ```w = 7206360 / 4657 x = 4893246 / 4657 y = 3515820 / 4657 z = 5439213 / 4657 ``` after reducing z to its lowest terms. So, for these four number to be positive integers, W, X, Y, and Z must be 4657 times the first column above. Thus, W = 2226 × 4657 = 10366482. And the rest of the numbers follow. ```W = 10,366,482 X = 7,460,514 Y = 7,358,060 Z = 4,149,387 w = 7,206,360 x = 4,893,246 y = 3,515,820 z = 5,439,213 ```

 Subject Author Date Answer K Sengupta 2008-11-16 00:10:42 why bother red_sox_fan_032003 2004-02-25 17:50:22 re: full solution - exact fractions Brian Smith 2004-02-18 11:32:13 re: Why bother? Penny 2004-02-18 04:36:50 Why bother? Ady TZIDON 2004-02-18 03:56:18 full solution SilverKnight 2004-02-17 21:28:47

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