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<P><BR>Let f(n) be the sum of all permutes of n.</P>
<P>Let</P>
<P> s(n) = sum of digits of n.<BR> d(n) =
number of digits of n.<BR> c_n(k) = number of occurences of
digit k in n.<BR> p(n) = PROD(k = 0..9;
c_n(k)!).<BR> r(n) = n-digit rep-1 number =
(10^n-1)/n.<BR> t(n) = ((s(n)(d(n)-1)!)/p(n)).</P>
<P>Then f(n) = t(n) r(d(n)).</P>
<P><BR>For example, if n = 314159, we get</P>
<P> s(n) = 23<BR> d(n) =
6<BR> c_n = (0, 2, 0, 1, 1, 1, 0, 0, 0,
1)<BR> p(n) = PROD(k = 0..9; c_n(k)!) =
2<BR> r(d(n)) = r(6) = 111111<BR> t(n) =
(23*120)/2 = 1380</P>
<P>and</P>
<P> f(314159) = 1380*11111 = 153333180</P>
<P><BR>A combinatorial argument shows that when we add the permutes of n, the
same number of each digit will appear in each column of the sum. This means
that</P>
<P> f(n) = (sum of each column) * (d(n)-digit rep-1
number)<BR> = (sum of each
column) * r(d(n))</P>
<P>from which we conclude</P>
<P> t(n) = sum of each column</P>
<P>implying that t(n) is integer and r(d(n)) | f(n).</P>
<P><BR>Using this knowledge, we can explore why f(n) = 29997 is insoluble while
f(n) = 2997 is not.</P>
<P>Let n have d(n) digits. Then n <= 9 r(d(n)) and has at most d(n)!
permutes. This means that</P>
<P> [1] f(n) <= 9 r(d(n)) d(n)!</P>
<P>Suppose f(n) = 29997. We know that r(d(n)) | f(n), and the only rep-1 numbers
that divide 29997 are r(1) = 1, r(2) = 11 and r(4) = 1111, implying d(n) = 1, 2
or 4. From [1],</P>
<P> d(n) = 1 => f(n) <=
9<BR> d(n) = 2 => f(n) <=
198<BR> d(n) = 4 => f(n) <= 239976</P>
<P>Given f(n) = 29997, we must have d(n) = 4 and r(d(n)) = 1111.</P>
<P>This means that</P>
<P> t(n) = f(n)/r(d(n)) = 29997/1111 = 27.</P>
<P>so that</P>
<P> ((s(n)(d(n)-1)!)/p(n)) = 27.</P>
<P>Since d(n) = 4, we have</P>
<P> 6 s(n)/p(n) = 27<BR> => s(n) =
9(p(n)/2).</P>
<P>Since p(n) and s(n) are both integer, p(n) must be even to make the right
side integer. This means 9 | s(n) <= 9 d(n) = 36. This means that s(n) = 9,
18, 27 or 36.</P>
<P>Suppose s(n) = 9. Then p(n) = 2. By the definition of p(n), p(n) = 2 iff
exactly one digit appears twice in n and all others once (or not at all). So we
are looking for a 4-digit n with a single repeated digit and digit sum 9. Many
such numbers exist, 1008 being the smallest, and indeed f(1008) = 29997.</P>
<P><BR>But now look at f(n) = 2997. Following similar logic to the above,
we find that d(n) = 3 and t(n) = 27 giving</P>
<P> ((s(n)(d(n)-1)!)/p(n)) = 27<BR> => 2
s(n)/p(n) = 27<BR> => s(n) = 27(p(n)/2).</P>
<P>Again, p(n) must be even, so 27 | s(n). But s(n) <= 9 d(n) = 27, so s(n) =
27 forcing n = 999 and f(n) = 999. Hence f(n) = 2997 is insoluble.</P>
<P> </P>
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