What is the Answer? See the additional details.?

For the discussion purpose, let the original three-digit number to be: 100a + 10b + c. We are given that:

0 < a < 10, 0<= b < a, 0 < c < a, and 10 + c - a = 4.

From 10 + c - a = 4, we know a - c = 6.

Therefore, the requested difference between the three-digit number and its everse is:

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

= 100(a - 1 -c) + 90 + 10 + c - a

= 594

There are not many such three-digit numbers which meet the given conditions. Since the minimum of c is 1, the minimum of a is 7. Also c is completely determined by a. Therefore, there are only 7 + 8 + 9 = 24 different such three-digit numbers.

1

According to your question

The Unit digit numbers should be only 3,2,1,0

The Hundred numbers should be only 9,8,7,6

Then Only the difference will be 4 in units place,and what ever the value we keep in the tens place(Which is lesser than hundreds value) the answer is same

i.e the difference value must be 594

594.

the number of combinations are

if the Hundreds place is 9 then there are -9 numbers

if the Hundreds place is 8 then there are -8 numbers

if the Hundreds place is 7 then there are -7 numbers

if the Hundreds place is 6 then there are -6 numbers

so total 9+8+7+6 = 30

So there are 30 numbers

Answers are 594,30

2

1131111112

2111111311

4

u cheat ... u give a condition dat will have -1 in reverse every time when subtracted :-p