SATHEE: Permutations and Combinations 2 Question 24

Permutations and Combinations 2 Question 24

24. $m n$ squares of equal size are arranged to form a rectangle of dimension $m$ by $n$ where $m$ and $n$ are natural numbers. Two squares will be called ’neighbours’ if they have exactly one common side. A natural number is written in each square such that the number in written any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the numbers used are equal.

(1982, 5M)

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Solution:

Let $m n$ squares of equal size are arrange to form a rectangle of dimension $m$ by $n$ . Shown as, from figure.

neighbours of $x_{1}$ are $x_{2}, x_{3}, x_{4}, x_{5} x_{5}$ are $x_{1}, x_{6}, x_{7}$ and $x_{7}$ are $x_{5}, x_{4}$ .

$\begin{array}{ll} \Rightarrow & x_{1} = \frac{x_{2} + x_{3} + x_{4} + x_{5}}{4}, x_{5} = \frac{x_{1} + x_{6} + x_{7}}{3} \\ and x_{7} = \frac{x_{4} + x_{5}}{2} \\ ∴ 4 x_{1} = x_{2} + x_{3} + x_{4} + \frac{x_{1} + x_{6} + x_{7}}{3} \\ \Rightarrow 12 x_{1} = 3 x_{2} + 3 x_{3} + 3 x_{4} + x_{1} + x_{6} + \frac{x_{4} + x_{5}}{2} \\ \Rightarrow 24 x_{1} = 6 x_{2} + 6 x_{3} + 6 x_{4} + 2 x_{1} + 2 x_{6} + x_{4} + x_{5} \\ \Rightarrow 22 x_{1} = 6 x_{2} + 6 x_{3} + 7 x_{4} + x_{5} + 2 x_{6} \end{array}$

where, $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are all the natural numbers and $x_{1}$ is linearly expressed as the sum of $x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ where sum of coefficients are equal only if, all observations are same.

$\Rightarrow x_{2} = x_{3} = x_{4} = x_{5} = x_{6}$

$\Rightarrow$ All the numbers used are equal

Permutations and Combinations 2 Question 24

Solution:

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Permutations and Combinations 2 Question 24

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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