Question

Find the largest natural number by which the difference between a three-digit number and the number obtained by reversing its digits is always divisible .

Answer 1

Step-by-step explanation:

• Let 100a+10b+c be your number; its reverse is as we know 100c+10b+a.
• as we subtract ,
• (100a+10b+c)−(100c+10b+a)=
• 100a - a + c - 100c + 10b - 10b
• (100−1)a+(10−10)b+(1−100)c =
• 99a−99c=
• 99(a−c)
• So it is always divisible by 99, and in particular, 9.
• pls do mark it as brainliest : )