Question

how many three digit number are there such that sum of the number and its reverse is divisible by 101​

Answer 1

Step-by-step explanation:

Let abc be the number.

Thus abc = 100a + 10b + c

The number obtained by reversing the number 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.