Question

The middle digit of a number between 100 and 1000 is zero, and the sum of the other digits is 11. If the digits are interchanged, the number so formed exceeds the original number by 495. Find the number.

Answer 1

let the three digits number be 100A+0B+C,=100A+C, whose middle digit is zero,

then According to question,

A+C=11......eq(1),

now,

(100C+A)-(100A-C)=495,

99C-99A=495,

then

C-A=5,........eq(2),

on solving eq(1) and eq(2), we get

C=8,

A=3,

therefore
number will be 100×3+8=300+8=308