[Guest]

Realised that my previous number reversal problem probably wasn't interesting enough at first sight to warrant any tough thinking. So I thought I'd show you guys this rather neat result, and then see if there's more interest in the harder puzzle:

1. Start with any 3 digit number, that isn't the same when written backwards (e.g. 123 is allowed, but 121 is not).

2. Write it backwards, and subtract the smaller from the larger (e.g. 321 - 123 = 198).

3. Take this number, reverse it again, but this time, add the reversed number (e.g. 198 + 891 = 1089).

Does this final answer always equal 1089? If so, prove it. If not, find an exception.

If you run into any 1 or 2 digit numbers along the way, treat them as having a leading zero. (e.g. think of "99" as "099", which, when reversed, would be written as 990. Similarly 100, when reversed, would be 001, or simply 1.)

[Guest]

Yeah this is such a cool math trick i have found why this happens here(Can't get it though! XD)

http://incrediblefacts.com/numbers-and-mathematics/1089-a-magic-number/

[araver]

Let N=abc=100*a+10*b+c, where a<>c (else N is palindromic).

Reverse(N)=cba=100*c+10*b+a.

F(N) = abs(N-Reverse(N)) > 0

R(N) = N + Reverse(N)

Note that

F(N)=F(Reverse(N))

so without loss of generality we can assume that N > Reverse(N) or equivalently that a>c. Then

F(N)=100*a+10*b+c-(100*c+10*b+a)=99*(a-c).

Since

 9 > a > c >= 0 

then a-c is one of the following:

{1,2,3,4,5,6,7,8,9}

. Therefore F(N) is a multiple of 99, one of the following:

{099, 198, 297, 396, 495, 594, 693, 792, 891}

. Note that the middle digit is always 9 and equal to the sum of the other 2 digits (including the leading zero when appropriate). Note that for 3-digit numbers

R(N)=N+Reverse(N)=100*a+10*b+c+100*c+10*b+a=101*(a+c)+20*b.

From the observation above for F(N) we get

 R(F(N))=101*9+20*9=1089.

Realised that my previous number reversal problem probably wasn't interesting enough at first sight to warrant any tough thinking. So I thought I'd show you guys this rather neat result, and then see if there's more interest in the harder puzzle:

It is indeed interesting, but I did not find a complete proof so far. Certain cases are easier to describe. Will post my thoughts soon

[Akriti]

Take any 3 digit number, abc. Writing the subtraction gives:

abc – cba = 100a -100c – a – c =99a – 99c = 99(a-c)

Since a and c differ by at least 2, the possible values of a-c are: 2,3,4,5,6,7,8

So, 99(a-c) must equal one of the following: 198, 297, 396, 495, 594, 693, 792

Now say that the result of abc – cba is def

Then the end result,P, of adding def to fed can be written as 100d + 10e + f + 100f + 10e +d

P = 100(d+f) + 20e +d + f

But from our list of possible values of 99(a-c), we can can see that the middle digit, e, is always 9. Also, the addition of the two end digits is always 9. So d+f = 9.

So our final sum P = 100(9) + 20(9) + 9

P= 900 + 180 + 9

P=1089

Edited November 16, 2010 by Akriti

[k-man]

It's always 1089. Here is the proof...

If you express the original 3-digit number as 100a + 10b + c, where a, b, c are the digits of the number, then the reversed number will be 100c + 10b + a. Subtracting one from the other we get

100a + 10b + c - 100c - 10b - a = 99a - 99c = 99*|a-c|, where |a-c| is an absolute value of a-c

a and c are different single digit numbers, so 1 <= |a-c| <= 9 and our result space is limited to

1*99 = 99

2*99 = 198

3*99 = 297

4*99 = 396

5*99 = 495

6*99 = 594

7*99 = 693

8*99 = 792

9*99 = 891

At this point you could enumerate all possible results of adding these numbers to their reverse and see that the result will always equal 1089. You could also see that all these numbers have a common property that the middle digit is 9 and the sum of the outside digits is also 9. Adding them to their reverse number will always equal to 900 + 180 + 9 = 1089