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

Answer:

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.

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all the best for coming exmz !1

Answer 2

Answer:

Step-by-step explanatory the 3 digit number be 100a+10b+c

Then ATQ

(100a+10b+c)-(100c+10b+a)

= 99a - 99c

And now we will take 99 as common

Then the ans in 99 (a-c)

=99