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There are some ten digit numbers which when reversed and processed in a special way, something like this happens....

e.g. The number 4204234125 in reverse is 5214324024, If we part the digits in the number 4204234125 as 4,20,42,3,41,25 and add them we get the sum which is equal to if we do the same with its reverse number 5214324024, namely 5,21,43,2,40,24, and add these numbers with each other. i.e. 4+20+42+3+41+25 = 5+21+43+2+40+24. Another such ten digit number is 1223343322.

How many such numbers exist? How to find them?

The partition must be like this: 1 digit, 2 digits, 2 digits, 1 digit, 2 digits, 2 digits.

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  • 2
    $\begingroup$ Which ways to partition them are allowed? $\endgroup$ – user480281 Oct 6 '17 at 5:06
  • $\begingroup$ Like this: 1 digit,2 digit,2 digit,1 digit,2 digit,2 digit $\endgroup$ – Deepeshkumar Oct 6 '17 at 5:10
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    $\begingroup$ @Deepeshkumar How many numbers have you tried? Have you found a single one that doesn't fit the pattern? $\endgroup$ – Erick Wong Oct 6 '17 at 5:13
  • $\begingroup$ Does $3,15,27,3,13,29$ , $3152731329$ Works ? $\endgroup$ – Ahmad Oct 6 '17 at 5:17
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Every ten digit number has this property.

Look at the digits in the partitioned sums in the example. The 4 ends up in the ones place of the 24. The 2 in the tens place of 20 ends up in the tens place of 24. The zero in the ones place of the 20 ends up in the ones place of the 40 and so on. Clearly the reversed partitioned sum would have the same value as the original partitioned sum.

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    $\begingroup$ a + (10b + c) + (10d + e) + f + (10g + h) + (10i + j) = j + (10i + h) + (10g + f) + e + (10d + c) + (10b + a) I think this makes it pretty obvious. Note that it also works in any other base. $\endgroup$ – Christoph Oct 6 '17 at 7:48
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    $\begingroup$ and presumably there is something similar for any number with $4n+2$ digits $\endgroup$ – Henry Oct 6 '17 at 9:02
  • $\begingroup$ Or at least every ten digit number that doesn't end in 0 (depending on how you interpret the rules). $\endgroup$ – cjm Oct 6 '17 at 19:01
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HINT.-There are a lot, because besides the examples like the given by the OP, we have also the palindromic numbers of ten digits $$abcdeedcba\text{ where five digits a,b,c,d,e }\text{ with } a\ne 0 $$ There are of these palindromic numbers$$9\cdot10\cdot10\cdot10\cdot10=90000$$

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