There are a total of 100 animals: cows, sheep and buffaloes. These 100 animals ate 100 bunches of grass.

Every cow ate 5 bunches, every buffalo ate 3 bunches and every sheep ate only 1/3 bunch.

How many cow, sheep and buffalo are there? You only know that there is at least one of every kind of animal.

Let the respective number of cows, buffaloes and sheep be a, b and c, giving:

a + b + c = 100

5a+ 3b + c/3 = 100

Or, 2b + 14c/3 = 400

0r, 3b + 7c = 600

Reducing both sides in mod 3;

7c = 0 (mod 3); so that c must be divisible by 3 as 3 and 7 are coprime to one another

Let c = 3d. This yields:

b = 200 - 7d

So, a+ b+ c = 100 gives:

a = 4d - 100

Since there must be at least one kind for wach, it follows that

each of a and b must be positive.

Now b> 0 gives d<200/7< 203/7 = 29

and a>0, gives d> 25

So d can assume values between 26 and 28 inclusively.

d = 26, gives (a, b, c) = (4, 18, 78)

d= 27, gives (a, b, c) = (8, 11, 81)

d = 28, gives (a, b, c) = (12, 4, 84)

So, (#cows, # buffaloes, # sheep) = (4, 18, 78); (8, 11, 81) ; (12, 4, 84)

*Edited on ***March 24, 2007, 12:06 pm**