Question

How can 90cows be tied on 13 pillars with an odd number on each pillars

Answer 1

Answer:

It can never be, it's impossible!

Concept used:

The sum of odd positive integers whose no. is also odd, is always odd.

Step-by-step explanation:

⇒ According to the question, each pillar should be tied with cows whose no. is odd each.

⇒ In respect to the concept given above, when considering any three pillars, the total no. of cows is an odd number. Like this, while considering five pillars, also here an odd number the total no. of cows. Thus, so is the total no. of cows in the all 13 pillars.

⇒ But it should be 90 cows as given in the question!!!

Method:

⇒ As odd numbers leave remainder 1 on division by 2, let the no. of cows in each pillar be 2a + 1, 2b + 1, 2c + 1, 2d + 1, 2e + 1, 2f + 1, 2g + 1, 2h + 1, 2i + 1, 2j + 1, 2k + 1, 2l + 1 and 2m + 1, for any whole numbers a, b, c, d, e, f, g, h, i, j, k, l, m.

⇒ Thus,

• 2a + 1 + 2b + 1 + 2c + 1 + 2d + 1 + 2e + 1 + 2f + 1 + 2g + 1 + 2h + 1 + 2i + 1 + 2j + 1 + 2k + 1 + 2l + 1 + 2m + 1 = 90
• 2a + 2b + 2c + 2d + 2e + 2f + 2g + 2h + 2i + 2j + 2k + 2l + 2m + 13 = 90
• 2(a + b + c + d + e + f + g + h + i + j + k + l + m) + 13 = 90
• 2(a + b + c + d + e + f + g + h + i + j + k + l + m) = 90 - 13
• 2(a + b + c + d + e + f + g + h + i + j + k + l + m) = 77

⇒ Here it contradicts that the LHS is even while the RHS is odd!!!

⇒ As the method continues,

• 2(a + b + c + d + e + f + g + h + i + j + k + l + m) = 77
• a + b + c + d + e + f + g + h + i + j + k + l + m = 77 / 2
• a + b + c + d + e + f + g + h + i + j + k + l + m = 38.5

⇒ Here, how a decimal number can be written as sum of some whole numbers???!!!