Question

How can 100 cows be tied on 9 pillars with an odd number on each pillar?

Answer 1

It is impossible to tie 100 cows on 9 pillars each with odd no. of cows on each pillar!!!

Because of the concept given below:

The sum of odd numbers, whose no. is also odd, is always odd!

This concept means that, as an example, the sum of three odd numbers, five odd numbers, seven odd numbers, etc. are also odd.

In this question, odd no. of cows are to be tied on 9 pillars each, where the total no. of cows is 100. As there are odd numbers on each 9 pillar, according to the concept, the sum of 9 odd numbers is also odd, but the total no. of cows 100 is even.

Let me explain it precisely, using algebra.

As odd numbers leave remainder 1 on division by 2, let the no. of cows in each cow be 2a + 1, 2b + 1, 2c + 1, 2d + 1, 2e + 1, 2f + 1, 2g + 1, 2h + 1, 2i + 1, where a, b, c, d, e, f, g, h and i are whole numbers.

Thus the sum of all these is 100.

  ⇒ 2a + 1 + 2b + 1 + 2c + 1 + 2d + 1 + 2e + 1 + 2f + 1 + 2g + 1 + 2h + 1 + 2i + 1 = 100

  ⇒ 2a + 2b + 2c + 2d + 2e + 2f + 2g + 2h + 2i + 9 = 100

  ⇒ 2(a + b + c + d + e + f + g + h + i) + 9 = 100

  ⇒ 2(a + b + c + d + e + f + g + h + i) = 100 - 9

  ⇒ 2(a + b + c + d + e + f + g + h + i) = 91

Here it creates a contradiction that the LHS is even while the RHS is odd.

As the process continues, we get,

  ⇒ (a + b + c + d + e + f + g + h + i) = 45.5

And it is known that no decimal numbers can be written as sum of two or more whole numbers.

Thus it is impossible to tie 100 cows on 9 pillars, with an odd number of cows on each pillar.

Answer 2

Answer:

I guess this wont be possible in a simple case. Below is the explanation.

Suppose the pillars from 1 to 9 have x1, x2, x3, x4, x5, x6, x7, x8 and x9 number of cows tied, where x1, x2...x9 are all odd numbers.

Now, accoring to question,

x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 100

in other words, odd + odd + odd + odd + odd + odd + odd + odd + odd = 100

=> (odd + odd) + (odd + odd) + (odd + odd) + (odd + odd) + odd = 100

=> [(even) + (even) + (even) + (even)] + odd = 100

=> Even + odd = 100.

But 100 is an even number.

Because Even + odd will never be an even number, hence the result will never be 100.

Since we are talking about real cows, there is no fractional number, no negative numbers etc.

But, since nothing is impossible, you can take this question as a riddle and tie one cow to two pillars simultaneously, using two ropes.

Another clever answer can be derived from the question itself. You mentioned "with an odd number on each pillar".

So you can take take 17 pillars and label them as Pillar 1, 2, 3....17. Then you can tie the cows only on Odd-numbered pillars