Question

the middle digit of a 3 digit number is 0 and the sum of other two digit is 11. if the number obtained by reversing the digits exceeds original number by 495. find the number​

Answer 1

Answer:

The number is 308

Step-by-step explanation:

Let us assume that the number is 'abc'

Given information : b = 0 and a + c = 11

A 3 digit number can be written as 100a + 10b + c

The reverse would be cba which is, 100c + 10b + a

From the question, we know that the abc = cba + 495

100a + 10b + c = 100c + 10b + a + 495

99a - 99c = 495

a + c = 11

We get a = 8 and c = 3

The number is 308 and 308+495 = 803