Answers only please?

Question

Write down any 3 digit number. It can be any number as long as the first and last digit differ by more than one. Now reverse the digits and take the smaller number away fron the larger one.

With your answer, now reverse these digits and add these two numbers together.

Please display your answer and compare it to other answers!

Answer 1

The answer is 1089. If the digits of the original number are denoted by x, y, and z, and assuming x>z, then the value of the original number is 100x+10Y+z, the value of the reversed number is x+10y+100z, and the difference is 99x-99z = 99(x-z). Let the decimal expansion of this be denoted by 100p+10q+r. The value of q is always 9. The value of the number reversed is 100r+90+p, so the sum is 101(p+r)+180. It can be shown that p+r = 9, so the sum is 909+180 = 1089.
-- Robert A. Saunders, Lake Stevens, WA.

Answer 2

The answer is always and in every possible conditon of things equal to 1089. Consider that a three-digit number abc can be represented algebraically as 100a + 10b + c. Assume without loss of generality that a>c. Then reversing the numbers and subtracting gives you 100a+10b+c-(100c+10b+a) ⇒ 99a - 99c ⇒ 99(a-c). Since a and c are both one digit numbers, their difference cannot be greater than 9 (given by a=9 and c=0). Further, the problem stipulates that the difference is greater than 1, ergo a-c must be an integer between two and nine (inclusive). The only possible values for the number after the first step are therefore 2*99, 3*99... 9*99, which are: 198, 297, 396, 495, 594, 693, 792, and 891. It may be verified manually that reversing any of these numbers and adding that to itself produces 1089.

Answer 3

3

Answer 4

1089

Answer 5

1089

Answer 6

The answer is always 1089. I tried with 10 different sets of numbers.

Answer 7

1089

Answer 8

1089

Answer 9

55

Answer 10

I got 1089