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Let the unit digit of original number be 'a'. So, tens digit of original number = a2 ATQ, 10a2 + a + 10a + a2 = 132 Or, 11(a2 + a) = 132 Or, a2 + a - 12 = 0 Or, a2 + 4a - 3a - 12 = 0 Or, a(a + 4) - 3(a + 4) = 0 Or, (a - 3) (a + 4) = 0 So, 'a' = 3 or 'a' = - 4 Since, the original number is natural. So, 'a' = 3 Therefore, required value = 0.65 X (10a2 + a - 10a - a2) = 0.65 X 9 X (32 - 3) = 0.65 X 54 = 35.1
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