$I f c > a; show that c b a - a b c = 99 (c - a) .$

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Solution

$G i v e n : c > a To show : c b a - a b c = 99 (c - a) P r o o f : c b a = 100 c + 10 b + a . . . (i) (By using property 3) a b c = 100 a + 10 b + c . . . (i i) (By using property 3) Subtracting (ii) from (i) c b a - a b c = 100 c + 10 b + a - 100 a - 10 b - c \Rightarrow c b a - a b c = 99 c - 99 a \Rightarrow c b a - a b c = 99 (c - a) Hence Proved.$

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