The sum of a 3 digit number a0b and the number formed by reversing its digit is always divisible by .

111

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101

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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.

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The sum of a three digit number and the number formed by the reversal of its digits (if for both the numbers middle digit is 0) is always divisible by 101

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The sum of a 3 digit number a0b and the number formed by reversing its digit is always divisible by .

The correct option is C 101The 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.

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.