po russkii, all is 3

<t>notice in the general case that1- there is at least one '0', otherwise a_i >= 1 for all i, sum_i a_i =n => a_i = 1 for all i, but 1....1 is not a fixed point of the transformation2- As before, let R = # of nonzero digits represented; then R = (n-1) - a_0, where a_0 = # of zeroes3- Then sum_i a_i ...

without listing the partitions for 8, we haveAVRMHOH...BM...CN...DPN...EO...FJP...GQN...II...KI...LUSM...RM...TM...I = 00001124 -> 4210100 as below,2cycle isO = 00000134 -> H = 51011000 -> 43000100 -> 51011000...

<t>for 7 digits, putA 1111111B 0111112C 0011113D 0011122E 0001114F 0001123G 0001222H 0000115I 0000124J 0000133K 0000223L 0000016M 0000025N 0000034O 0000007Then we have AOLHIEJI... / BH... / CI... / DKI... / FF... / GJ... / MH... / NH...so the only solutions are F (1cycle) and IEJ (3cycle)Upon liftin...

<t>for 6 digits, there is no solution. PutA = 111111B = 011112C = 001122D = 000222E = 001113F = 000123G = 000033H = 000024I = 000015J = 000006K = 000114(maybe this isn't the standard way, I just wrote them down as they came to my head)Then we see that we haveAJIKFEFEFEFEFE.... i.e. ends in a 2-cycle...

<t>to add a little color, the quotient space is just the space of partitions of n, where the mapping just zero-pads the partitionso for example, k+1 = 4, partitions are 1111 0112 0022 0013 0004 (as before, sum of a_i's = 4 as well, might as well restrict right away and notice that this space is pres...

<t>what I realized was that the solution is a fixed point, so if you start with any sequence a_0a_1a_2..a_k, and putf(A) = (#0s in A)(#1s in A)...(#ks in A)then a solution to the problem has to be a fixed point of f. There is a problem in that all the numbers could be the same, in which you'd have k...

fwiw:1210, 202021200--3211000 / 42101000 / 521001000 / 6210001000 / 72100001000etc.

in order to make the problem more interesting you may trya^(mb) = b^(na)for instance with (m,n) = (1,2), we have the solution (3^2, 3^3) i.e. both sides are 3^54and with certain combinations of m and n, a new family of solutions can appear

<t>write c = (r/s)^q, where q is maximal, r and s have no common factors, (a,b) = ( c^(1 / (1-c) ), c^(c / (1-c) ) )consider the situation where (r^q - s^q) divides q.s^q; s < r and you will see that q is very limited as a function of r - in particular, exp(3/2) < 5 so q <= 4, and in nearly all case...

yes, but they are /Q... indeed I challenge you to prove that all rational solutions are of this formnow there might be some solutions over Q(sqrt(p))... haven't really looked into that

hehehe my solutions of a^b = b^a are(9/4, 27/8)(625/256, 3125/1024)(7776/3125, 46656/15625)...but anybody can figure out the pattern here

<t>1) a <> 02) Let R = # of non-zero digits represented. I.e. if the number is 12320, R = 3. Then R = 4 - {# of b,c,d,e which are zero}, by definition of b,c,d,e. But since a<>0, R = 4 - {# of zeroes in the number} = 4 - a.3) a+b+c+d+e = 5, b/c there are 5 digits in the number.4) Now we rule out the...

if you go towww.losers.comI think you'll see that they already have you on their e-mail list

vanilla 'elephant in the living room' situationbut the whole thing's purely rhetoricalthere is only one definition of squandered opportunity: squandered opportunitywhy ask why?