Question
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
Answer: Option A
:
A
Let abc be the number.
Thus abc = 100a + 10b + c
The number obtained by reversing the digits is cba.
cba = 100c + 10b + a
Adding the two numbers we get, abc + cba = (100a + 10b + c) + (100c + 10b + a) = 101a + 20b + 101c
Since middle digit is 0, the sum will be 101(a+c), which is divisible by 101.
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:
A
Let abc be the number.
Thus abc = 100a + 10b + c
The number obtained by reversing the digits is cba.
cba = 100c + 10b + a
Adding the two numbers we get, abc + cba = (100a + 10b + c) + (100c + 10b + a) = 101a + 20b + 101c
Since middle digit is 0, the sum will be 101(a+c), which is divisible by 101.
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