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Given a number of the form $0.xyzxyzxyz$, where $x,y,z$ are distinct integers taking on values $\in \{0,1,2, \ldots, 9\}$, what is the sum, $S$, of all such numbers of the form $0.xyzxyzxyz$?

Here is how I solved this problem. For each $x$, there are 72 corresponding $xyzxyzxyz$ values. So each integer for $x$ will occur 72 times and similarly for $y$ and $z$. $72 * \sum_{i=0}^9 i = 3240$.

So the summation ends up looking something like $$ 324.000000000 + 32.400000000 + 3.240000000 + \ldots + 0.000003240 \\ \approx 360 $$

The answer that was given said it is $0.111111111 * 72 * \sum_{i=0}^9 i$. Where did the $0.111111111$ come from? This approach is simpler than mine, but I just don't know where this decimal came from.


So it appears that the approach is very similar to mine, but they approached it from the form $0.1 * 3240 + 0.01 * 3240 + 0.001 * 3240 + \ldots = 0.000000001 * 3240 = (0.1 + 0.01 + \ldots + 0.000000001) * 3240 = 0.111111111 * 3240$

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  • $\begingroup$ For every block $xyz$ define the "conjugate" block $\overline {xyz}$ by replacing $x$ by $9-x$ and so on. Thus $\overline {391}=608$, for example. Then $.xyzxyzxyz+.\overline {xyzxyzxyz}=.999999999$. Now there are $5\times 9\times 8$ ways to choose $xyz$ with $x≤4$ so the answer is $.999999999\times 5\times 9\times 8$. That's at least similar to their method, $\endgroup$ – lulu Aug 30 '20 at 22:26
  • $\begingroup$ Note: it's not clear (to me) whether you intended your decimal to repeat indefinitely. If you did, then of course my string $.999\cdots$ is just $1$. $\endgroup$ – lulu Aug 30 '20 at 22:29
  • $\begingroup$ Oh, good call. It is certainly possible OP meant for this to be a terminating decimal. It doesn't change the analysis of the situation much in any event... just trim the decimals as necessary. $\endgroup$ – JMoravitz Aug 30 '20 at 22:33
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The digit $1$ will occur in the $x$ positions precisely $\frac{1}{10}$ of the time. Similarly it will occur in the $y$ positions precisely $\frac{1}{10}$ of the time and same to for the $z$ positions.

The contribution of the $1$'s digit to the overall sum from a single term in the sum when it occurred in the $x$ position would be $0.100100100100\cdots$. Similarly in the $y$ position it would have contributed $0.010010010010\cdots$ and for the $z$ position as $0.001001001\cdots$. Note again that these each occur just as frequently as one another.

As there are $10\times 9\times 8$ ways to fill in the $x$'s, $y$'s and $z$'s such that they are all different digits, the total contribution of all of the $1$'s in the final summation who occurred in the $x$ position will be $\frac{1}{10}\times 10\times 9\times 8\times 0.100100100\cdots$ and in the $y$ position will be $\frac{1}{10}\times 10\times 9\times 8\times 0.010010010\cdots$ and similarly for the $z$ position.

Adding these together gives the total contribution of all $1$'s in the summation as $72\times 0.1111111\cdots$

The same argument applies for the total contribution of the $2$'s and $3$'s etc... yielding the final result

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