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Question

Problem Of the Week Suppose the hundreds digit of a three-digit number is greater than the tens digit and the ones digit. When the digits in the number are reversed and the resulting number is subtracted from the original number. The units digit in the difference is 4. What is the difference between the three-digit number and its everse? How many different three-digit numbers meet the given conditions?

What is the difference between the three-digit number and its *Reverse*? How many differents three-digit numbers meet the given conditions?

Answer 1

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.

Answer 2

1

Answer 3

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

Answer 4

2

Answer 5

1131111112
2111111311

4

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