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

Posted by K Sengupta on 2007-03-08 12:08:00