Question

Difference of three digit and reverse is multiple of 99

Answer 1

Answer:

Let the number be 100a+10b+c (The unit place is c, tenth place is b and the hundredth place is a)

When it reverses, it is 100c+10b+a

Now 100a+10b+c-99 = 100c+10b+a

=>     99a - 99c = 99

=>     a - c = 1

Again, sum of digits are 9, so a+b+c = 9

For the number to be the smallest among the possible, a has to be least.

a can't be 0, and 1 (c will be -1 and 0, which are not allowed)

So a is 2

c = a-1 = 1

b = 9-(a+c) = 9-3 = 6

So the number is 261

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