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_____.

Open in App

Solution

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.

Suggest Corrections

Similar questions

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

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

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

Q. The sum of a two digit number and the number obtained by reversing its digits is always divisible by

Q. The difference between a 3 digit number and the number formed by reversing its digits is not always divisible by

Join BYJU'S Learning Program

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_____.

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.