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The sum of a three digit number (ABC) and another three digit number (DEF), where DEF is obtained by thoroughly shuffling* the original three digits of ABC, is 1000. Then, find all possible such pairs of numbers (ABC and DEF).

  • thoroughly shuffling means - after shuffling, no digit of the original number remains in its place!
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This is

not possible (well, depending on the definition, see below). DEF must be either BCA or CAB. Without loss of generality, we can assume it's BCA (otherwise, we can reverse the order of the summands).
_ABC
_BCA
---- +
1000
We see that $A+C$ must be $10$ (*) (if it's 0, $A = 0$, so ABC isn't a three digit number), carrying the 1 we see that $B+C+1=10$ and so $A+B+1=10$. Subtracting the latter two equations gives $A=C$, so by (*) they must both be 5 and so $B=4$.
So the only solution is 455+545=1000 and this is valid if you interpret the 'shuffle' as the 1's digit ending up as the 100's digit, and the 10's digit as the 1's digit.

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    $\begingroup$ FWIW, I have verified this via brute force. The only two allowable values for ABC were 455 and 545. $\endgroup$ – Engineer Toast May 10 '18 at 12:45
  • $\begingroup$ Yes, shuffle 's purpose (explicitly mentioned) is that. Good work @Glorfindel $\endgroup$ – Mea Culpa Nay May 10 '18 at 14:41
  • $\begingroup$ If ABC were 455, DEF could be forrmed as CAB. The one's place after the shuffle might have the same value as before, that wouldn't mean it couldn't be the five that had been in the tens place (with the five that had been in the one's place moving to the hundreds). $\endgroup$ – supercat May 10 '18 at 21:47
  • $\begingroup$ "We see that A+C must be 10" - you should probably show why it can't be 0 as well. $\endgroup$ – Chris May 11 '18 at 9:21
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I'll come at this from a slightly different perspective:

The only two derangements of a set of three digits are their cyclic permutations. This means that two numbers in question are $100 A + 10 B + C$ and $100 B + 10 C + A$. We therefore have $$101A + 110 B + 11 C = 1000,$$ which we can rearrange to $$(100A + 110 B + 10C) + A + C = 1000.$$
Since the right-hand side is divisible by 10, and the terms in brackets on the left-hand side are automatically divisible by 10, we have must have $A + C$ divisible by 10 as well; and since they are single digits, we conclude that $A + C = 10$ or $A = C = 0$. In the latter case, we have $110B = 1000$, or $11B = 100$; but 100 is not divisible by 11, so this is impossible. Thus, $A + C = 10$.

Substituting this in, we obtain $101 A + 110 B + 11(10 - A) = 1000$, which reduces to $$9A + 11 B = 89$$

Since 9 and 11 are relatively prime, a solution can be obtained by repeatedly subtracting 9 from 89 until we get a multiple of 11; we have $89 - 5(9) = 44$, and so a solution is $A = 5$, $B = 4$. All other solutions are then of the form $A = 5 + 11 m$, $B = 4 - 9 m$ for some integer $m$; but none of these solutions are single-digit numbers.

Thus, the only solution is $$A = C = 5, \quad B = 4$$

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    $\begingroup$ Michael - I've taken the liberty of formatting your answer with whitespace, to improve its readability, and reduce the wall-of-text effect. Feel free to amend/revert it if required. $\endgroup$ – Phylyp May 10 '18 at 13:59
  • $\begingroup$ Good interpretation, thanks. But I have a doubt- why /how the left-hand side part of the first equation is divisible by 10? Is it an assumption? Please clarify. $\endgroup$ – Mea Culpa Nay May 10 '18 at 14:39
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    $\begingroup$ @MeaCulpaNay: Sorry, I meant the right-hand side when I said that. I've edited the answer to clarify. $\endgroup$ – Michael Seifert May 10 '18 at 14:43
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    $\begingroup$ I don't think you have actually excluded the possibility of A = C = 0. Its not valid but you just assumed that rather than proved it. $\endgroup$ – Chris May 11 '18 at 9:20
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    $\begingroup$ @Chris: Good point. Fixed. $\endgroup$ – Michael Seifert May 11 '18 at 12:39
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This is a huge stretch, but you can do

$950 + 50.9 = 1000.9$ - which I know is not $1000$ but is really close :)

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