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a question about permutation in the digits in the decimal system

I dont get how it works but the method is:

think of a three didgit number with none of the same didgits ; 124 Reverse it ; 421 then take the smallest from the biggest ; 421-124=297 Then reverse your answer ; 792 Then add them together ; 792+297=1089

Now remember that number then do it again with a different number

867 768 867-768=099 990 990+99=1089

I dont get how it works

can you find a number that doesn't ???

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marked as duplicate by William, Micah, Henning Makholm, Ross Millikan, rschwieb Sep 5 '12 at 20:18

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See for example here: math.stackexchange.com/questions/183661/… –  martini Sep 5 '12 at 18:59
    
it is so magical ;) –  Seyhmus Güngören Sep 5 '12 at 19:20
    
Now generalize it. What happens if you do the same problem with, say, appropriate four-digit numbers? –  Rick Decker Sep 5 '12 at 20:02
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1 Answer

Let the digits be $\rm\:abc,\:$ wlog $\rm\,a > c.\,$ So it is $\rm\:100\,a+10\,b+c.\:$ Put $\rm\,d = a-c.\:$ Note $\rm\:0<d<10.$ $$\rm\begin{eqnarray} &&\rm 100\, a &+&\rm 10\, b &+&\rm c &&\\ -\ \ \ &&\rm 100\, c &+&\rm 10\, b &+&\rm a &&\\ \hline \\ =\ \ &&\rm 100\, d &&\rm &-&\rm d &&\rm \end{eqnarray}$$

In decimal, we have $\rm\ \ 100\,d\, -\, d\ =\ 100\ (\ d\,-\,1)\ +\ 90\ +\ (10-d),\:$ by borrowing. Thus

$$\begin{eqnarray} &&\rm 100(\ \ \ d\,-\,1) &+&\ \rm 90 &+&\rm \: (10-d) \\ +\ \ \ &&\rm 100(-d+10) &+&\ \rm 90 &+&\rm (-1+d) \\ \hline \\ =\ \ &&\rm 100\cdot 10 &+&\ \rm 80 &+&\rm \ 9 \end{eqnarray}$$

Note a carry into the hundreds digit $\: 100 + 80\, =\, 90 + 90.$

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1  
That seems more an answer than a hint =)! –  Pedro Tamaroff Sep 5 '12 at 19:33
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