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

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Solution

$G i v e n, a > c To show : abc-cba=99(a-c) Proof : a b c = 100 a + 10 b + c . . . (i) By using property 3 c b a = 100 c + 10 b + a . . . (i i) By using property 3 Subtracting, equation (ii) from (i), we get a b c - c b a = 100 a + c - 100 c - a a b c - c b a = 99 a - 99 c ∴ a b c - c b a = 99 (a - c) Hence proved.$

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