Taking the 3 Combinations of 3 Digit Numbers and Adding Them
The sum of a ...
Question
The sum of a 3 digit number a0b and the number formed by reversing its digit is always divisible by .
A
111
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B
99
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C
101
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D
11
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Solution
The correct option is C 101 The expanded form of a0b = 100a + 10 x 0 + b = 100a + b The expanded form of b0a = 100b + 10 x 0 + a = 100b + a Adding the two numbers we get, 100a + b + 100b + a = 101a + 101b = 101(a + b) which is divisible by 101.