Consider this a Christmas puzzle. Take composite 52 with all its prime factors 2,2,13 and rearrange them to form the number 2132, which equals $41*52$. Any multidigit prime factor remains unchanged: 13 stays 13 and 8191 stays 8191. Thus 49146 with ALL its prime factors being 2,3,8191 could be rearranged into 819132 or 281913 or 238191, and so on. Do you think there are many numbers like composite 52 which can divide a rearrangement of ALL its prime factors?
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1$\begingroup$ Why are multi-digit prime factors immune to rearrangement? Is there something that makes the problem trivial if we relax that assumption? $\endgroup$– John DoumaDec 21, 2018 at 18:56
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$\begingroup$ "Do you think there are many numbers like composite 52 which can divide a rearrangement of ALL its prime factors?" -> No proof here, but I bet there are infinitely many $\endgroup$– Bram28Dec 21, 2018 at 19:02
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$\begingroup$ By rearranging the digits in a multidigits prime you will get many more arrangements, hence, making the puzzle NOT a Xmas puzzle. The point is to keep it pure by having these primes being left UNdisturbed. The puzzle is, of course, trivial if primes are used. $\endgroup$– J. M. BergotDec 21, 2018 at 19:03
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$\begingroup$ $24=2^3\cdot3$, $2232=24\cdot93$. $44=2\cdot2\cdot11$, $2112=44\cdot48$. $\endgroup$– Gerry MyersonDec 21, 2018 at 19:24
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$\begingroup$ Do you want to toss in MORE results to OEIS for their Xmas present? $\endgroup$– J. M. BergotDec 21, 2018 at 20:00
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2 Answers
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There are lots of examples of these. Here's a multi-arrangement example.
$3168=2^5\cdot3^2\cdot11$
$3168\times35424=112223232$
$3168\times66674=211223232$
$3168\times69799=221123232$
$3168\times70174=222311232$
$3168\times73299=232211232$
$3168\times73334=232322112$
Complete list for integers up to $10000$:
N M M/N [prime factors of N]
24 2232 93 [2, 2, 3, 2]
44 2112 48 [2, 11, 2]
52 2132 41 [2, 13, 2]
105 735 7 [7, 3, 5]
114 3192 28 [3, 19, 2]
152 22192 146 [2, 2, 19, 2]
176 222112 1262 [2, 2, 2, 11, 2]
348 29232 84 [29, 2, 3, 2]
378 73332 194 [7, 3, 3, 3, 2]
474 3792 8 [3, 79, 2]
548 21372 39 [2, 137, 2]
576 22223232 38582 [2, 2, 2, 2, 3, 2, 3, 2]
612 233172 381 [2, 3, 3, 17, 2]
636 23532 37 [2, 3, 53, 2]
1518 112332 74 [11, 23, 3, 2]
1908 532332 279 [53, 2, 3, 3, 2]
1911 13377 7 [13, 3, 7, 7]
2688 223222272 83044 [2, 2, 3, 2, 2, 2, 2, 7, 2]
3168 222311232 70174 [2, 2, 2, 3, 11, 2, 3, 2]
3168 221123232 69799 [2, 2, 11, 2, 3, 2, 3, 2]
3168 232211232 73299 [2, 3, 2, 2, 11, 2, 3, 2]
3168 232322112 73334 [2, 3, 2, 3, 2, 2, 11, 2]
3168 211223232 66674 [2, 11, 2, 2, 3, 2, 3, 2]
3168 112223232 35424 [11, 2, 2, 2, 3, 2, 3, 2]
3204 233892 73 [2, 3, 3, 89, 2]
3425 51375 15 [5, 137, 5]
3905 11715 3 [11, 71, 5]
4704 23722272 5043 [2, 3, 7, 2, 2, 2, 7, 2]
5292 2333772 441 [2, 3, 3, 3, 7, 7, 2]
5824 132222272 22703 [13, 2, 2, 2, 2, 2, 7, 2]
6372 5932332 931 [59, 3, 2, 3, 3, 2]
7695 3331935 433 [3, 3, 3, 19, 3, 5]
7824 23221632 2968 [2, 3, 2, 2, 163, 2]
7868 228172 29 [2, 281, 7, 2]
7928 229912 29 [2, 2, 991, 2]
8064 2323222272 288098 [2, 3, 2, 3, 2, 2, 2, 2, 7, 2]
8064 7222223232 895613 [7, 2, 2, 2, 2, 2, 3, 2, 3, 2]
8208 232319232 28304 [2, 3, 2, 3, 19, 2, 3, 2]
8352 222923232 26691 [2, 2, 29, 2, 3, 2, 3, 2]
8398 1713192 204 [17, 13, 19, 2]
9072 322337232 35531 [3, 2, 2, 3, 3, 7, 2, 3, 2]
9072 372233232 41031 [3, 7, 2, 2, 3, 3, 2, 3, 2]
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$\begingroup$ Great present! Do you want to send it into OEIS? $\endgroup$ Dec 21, 2018 at 21:07
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$\begingroup$ I see just three odd numbers in this list: 105, 3425, and 3905. Perhaps there are odd numbers not ending in digit 5? $\endgroup$ Dec 22, 2018 at 18:13
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$\begingroup$ You don't think this answer is worth an upvote, J. M.? $\endgroup$ Dec 23, 2018 at 4:35
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$\begingroup$ To Myerson: I once added an positive vote and the machne yelled at me for violating some rule. I'll try it again to see what happens. Sure enough, it yelled again about voting for your own post. Can you vote for me..a proxy, of sorts? Do not forget that OEIS always likes Christmas presents in the form of sequences. That 3168 is certainly a spiritual number. $\endgroup$ Dec 23, 2018 at 21:51
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1$\begingroup$ OK, I'll try @Gerry and see what happens. I clicked at the top for original question and was promptly told not to do this. I took your suggestion and clicked nickgard answer and that made it 2 positive votes. For odd numbers, can there be one that does not end in digit 5? $\endgroup$ Dec 27, 2018 at 18:28
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Let $n$ be a natural number which can divide some concatenation of its prime divisors (with multiplicity, e.g. $52=2^2\cdot13$ has $2132=41\cdot 52$ as a concatenation of its prime factors), with the convention that the empty concatenation is $1$.
I am now scrambling to fix my arguments, please hold...
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$\begingroup$ I think the point about 52 is that it divides a rearrangement of all of its prime factors – it wasn't claimed to divide all rearrangements of its prime factors. $\endgroup$ Dec 21, 2018 at 19:20
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$\begingroup$ @GerryMyerson Whoops! Have you heard that reading comprehension is on the decline in America? $\endgroup$ Dec 21, 2018 at 19:21
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$\begingroup$ G Myerson is correct. The puzzle is to find just ONE of the many rearrangement of ALL prime factors that is divisible by n itself. To find some n with MORE than one divisibility of its many rearrangements would be a multiholiday puzzle. $\endgroup$ Dec 21, 2018 at 19:25