Consider a three-digit number N = 100x + 10y | vedclass

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Question

(D) Let the three-digit number be $N = 100x + 10y + z$.From property $I$: $(100x + 10z + y) - (100x + 10y + z) = 36$$\Rightarrow 9z - 9y = 36$ $\Right

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decreases by $540$

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(D) Let the three-digit number be $N = 100x + 10y + z$.From property $I$: $(100x + 10z + y) - (100x + 10y + z) = 36$$\Rightarrow 9z - 9y = 36$ $\Rightarrow z - y = 4$ (Equation $1$)From property $II$: $(100x + 10y + z) - (100z + 10y + x) = 198$$\Rightarrow 99x - 99z = 198$ $\Rightarrow x - z = 2$ (Equation $2$)Adding Equation $1$ and Equation $2$: $(x - z) + (z - y) = 2 + 4 \Rightarrow x - y = 6$.Now,we want to find the change when tens and hundreds digits are interchanged:New number $N' = 100y + 10x + z$.Change $= N - N' = (100x + 10y + z) - (100y + 10x + z) = 90x - 90y = 90(x - y)$.Substituting $x - y = 6$: Change $= 90 	imes 6 = 540$.Since $N - N' = 540$,the number decreases by $540$.

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