Base Ten Digit Sets
Alexm
as sets of single digits (without regard to order) then the two sets
are equal and equal to the set 24578. They have equal digit sets.
Another example is integers 5024 and 5025. Their cubes have equal
digit sets and are equal to the set 01223456688. Can you find another
integer pair n and n+1, n > 0, such that their cubes have equal digit
sets?
alexm
Mensanator
> 35^3 = 42875 and 38^3 = 54872. Looking at the digits of these cubes
> as sets of single digits (without regard to order) then the two sets
> are equal and equal to the set 24578. They have equal digit sets.
> Another example is integers 5024 and 5025. Their cubes have equal
> digit sets and are equal to the set 01223456688. Can you find another
Another? You would be better off finding a first one:
>>> digitsets(5024)
n**3 126808653824
(n+1)**3 126884390625
s0 digitset ['0', '1', '2', '2', '3', '4', '5', '6', '6', '8', '8',
'8']
s1 digitset ['0', '1', '2', '2', '3', '4', '5', '6', '6', '8', '8',
'9']
Oops! Try using your toes next time.
Alexm
>
> - Show quoted text -
Dittoopps! Well, does such a pair exist (using toes or fingers)?
Alexm
Mensanator
None less than n=10000000. You might have better luck if not
restricted
to adjacent integers such as you first example. But that's another
program.
>
> Alexm
Alexm
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
It looks unlikely that n and n+1 will work - by your result. Thanks.
Alexm
Chip Eastham
If n^3 and (n+1)^3 have the same digits
(up to rearrangement), then these would
be congruent mod 9, hence congruent mod
3 in particular.
But (n+1)^3 - n^3 = 3n^2 + 3n + 1.
regards, chip
me13013
> You might have better luck if not restricted to adjacent
E.g. 5^3 and 8^3.
Some others: 355^3 and 427^3; 656^3 and 674^3; 3772^3 and 4036^3;
5852^3 and 6470^3; 8744^3 and 8864^3. There are many more.
There are some cheats, like 1001^3, 1010^3 and 1100^3. OR 1002^3,
1020^3, 2001^3 and 2010^3.
I think more interesting than adjacent integers (which Chip has showed
impossible) would be finding pairs that are far apart (farthest I find
for 4 digits is 9938^3 and 4661^3), or that have a simple ratio. For
example, are there any others that have the ration 8/5?
Or how many are 3 apart, like these: 1095^3 and 1098^3; 1991^3 and
1994^3; 3714^3 and 3717^3; 4173^3 and 4176^3; 4823^3 and 4826^3;
6473^3 and 6476^3.
Bob H
Chip Eastham
I noticed that one or more of Bob's examples
of 3 apart pairs has both numbers divisible
by 3, e.g. 1095^3 and 1098^3.
Thus the cubes 365^3 = 48627125 and 366^3 =
49027896 have different digit sets, but on
multiplication by 27 the digit sets become
equal.
This suggests a more general question: If
m,n are positive integers, when does k exist
s.t. mk and nk have equal digit sets?
regards, chip
Mensanator
Interesting.
Especially m=855 and n=236.
838 571 414
['2', '3', '3', '4', '6', '9']
['2', '3', '3', '4', '6', '9']
718 536 9261
['3', '4', '6', '6', '8', '9', '9']
['3', '4', '6', '6', '8', '9', '9']
291 205 2772
['0', '2', '5', '6', '6', '8']
['0', '2', '5', '6', '6', '8']
811 605 10179
['1', '2', '5', '5', '6', '8', '9']
['1', '2', '5', '5', '6', '8', '9']
370 414 675
['0', '2', '4', '5', '7', '9']
['0', '2', '4', '5', '7', '9']
342 570 321
['0', '1', '2', '7', '8', '9']
['0', '1', '2', '7', '8', '9']
855 236 44658
['0', '1', '2', '3', '5', '8', '8', '9']
['0', '1', '2', '3', '5', '8', '8', '9']
374 171 6192
['0', '1', '2', '3', '5', '8', '8']
['0', '1', '2', '3', '5', '8', '8']
275 446 46
['0', '1', '2', '5', '6']
['0', '1', '2', '5', '6']
588 447 2607
['1', '1', '2', '3', '5', '6', '9']
['1', '1', '2', '3', '5', '6', '9']
995 320 484
['0', '1', '4', '5', '8', '8']
['0', '1', '4', '5', '8', '8']
267 672 1
['2', '6', '7']
['2', '6', '7']
150 469 11493
['0', '1', '2', '3', '5', '7', '9']
['0', '1', '2', '3', '5', '7', '9']
744 923 7893
['2', '2', '3', '5', '7', '8', '9']
['2', '2', '3', '5', '7', '8', '9']
427 121 12
['1', '2', '4', '5']
['1', '2', '4', '5']
780 524 585
['0', '0', '3', '4', '5', '6']
['0', '0', '3', '4', '5', '6']
332 837 99
['2', '3', '6', '8', '8']
['2', '3', '6', '8', '8']
986 710 426
['0', '0', '2', '3', '4', '6']
['0', '0', '2', '3', '4', '6']
584 710 84
['0', '4', '5', '6', '9']
['0', '4', '5', '6', '9']
526 575 999
['2', '4', '4', '5', '5', '7']
['2', '4', '4', '5', '5', '7']
>
> regards, chip
Chip Eastham
[snip further examples]
Consider m = 1 and n = 2. One solution
for k is 241280512605, but this is not
minimal. What is the smallest k such
that k and 2k have equal digit sets?
It would seem that for m = 1 and n = 10
no k can possibly give equal digit sets.
However if we include leading zeros,
solutions are possible. Is there a pair
m,n > 0 for which no solution k exists?
regards, chip
Chip Eastham
> Consider m = 1 and n = 2. One solution
> for k is 241280512605, but this is not
> minimal. What is the smallest k such
> that k and 2k have equal digit sets?
Hint: The periodic decimal expansion of a
familiar rational provides integer k s.t.
k, 2k, 3k, 4k, 5k, and 6k all have equal
digit sets.
--c
me13013
If we're going to allow 3k and 4k I will ignore the answer you're
thinking of and say 153846. It is not the smallest but it's not far
behind.
Or I could go with 117647058823529 which works for 2k, 3k, ... 8k. Or
105263157894736842 (2k .. 9k). Or
1016949152542372881355932203389830508474576271186440677966.
Bob H
Chip Eastham
> On Nov 9, 8:59 am, Chip Eastham <hardm...@gmail.com> wrote:
>
> > On Nov 7, 7:09 am, Chip Eastham <hardm...@gmail.com> wrote:
>
> > > Consider m = 1 and n = 2. One solution
> > > for k is 241280512605, but this is not
> > > minimal. What is the smallest k such
> > > that k and 2k have equal digit sets?
>
> > Hint: The periodic decimal expansion of a
> > familiar rational provides integer k s.t.
> > k, 2k, 3k, 4k, 5k, and 6k all have equal
> > digit sets.
>
> > --c
>
> If we're going to allow 3k and 4k I will
> ignore the answer you're thinking of and
> say 153846. It is not the smallest but
> it's not far behind.
But k = 153846 has a different digit set
than 2k has.
> Or I could go with 117647058823529 which
> works for 2k, 3k, ... 8k. Or
If think you meant 1176470588235294.
> 105263157894736842 (2k .. 9k). Or
> 1016949152542372881355932203389830508474576271186440677966.
It's like you found a mysterious crank to
turn!!! ;-)
If a leading zero is allowed on k, we can
extend the pattern to 10k, 11k, etc. E.g.
k = 0588235294117647, then:
2k = 1176470588235294,
...
10k = 5882352941176470.
11k = 6470588235294117,
...
16k = 9411764705882352.
If it were known that 10 is primitive for
Z/pZ with arbitrarily large primes p, and
allowing leading zeroes on k, then for all
integer m,n > 0 there exists k s.t. mk and
nk have equal digit sets.
However this remains an open question:
[Artin's conjecture on primitive roots -- Wikipedia]
http://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots
regards, chip
me13013
> On Nov 9, 7:28 pm, me13013 <me13...@gmail.com> wrote:
> > If we're going to allow 3k and 4k I will
> > ignore the answer you're thinking of and
> > say 153846. It is not the smallest but
> > it's not far behind.
>
> But k = 153846 has a different digit set than 2k has.
Yep. I wasn't clear, but I meant "only" 3k and 4k.
> It's like you found a mysterious crank to turn!!! ;-)
Some would argue that *I* am a crank.
Of course this kind of number is familiar to many of us. Because of
that, I think it would be more interesting to *exclude* numbers whose
multiple is simply a cyclic permutation of itself. You demonstrated k
and 2k for k=241280512605, which is not cyclic. That is more
interesting, in my opinion.
Bob H
me13013
> Of course this kind of number is familiar to many of us. Because of
> that, I think it would be more interesting to *exclude* numbers whose
> multiple is simply a cyclic permutation of itself. You demonstrated k
> and 2k for k=241280512605, which is not cyclic. That is more
> interesting, in my opinion.
Trying my hand at that, I found a non-cyclic number using each digit
0-9 exactly once in k and 2k: 1934708526. Is this hard to do? I
can't see any obvious process to generate these.
Bob H
g.r...@iit.cnr.it
> Trying my hand at that, I found a non-cyclic number using each digit
> 0-9 exactly once in k and 2k: 1934708526. Is this hard to do? I
> can't see any obvious process to generate these.
Among the 3,265,920 pandigital numbers p (here pandigital in the
strict sense, that is numbers composed by 10 distinct digits),
exactly 184,320 have the property that 2*p is pandigital as well,
the smallest being
1023456789 x 2 = 2046913578 and the largest
4938271605 x 2 = 9876543210.
None of these 184,320 numbers have 2*p as a cyclic permutation of p.
I have summarized the search results for all the multiples
(k=2,...,9)
of pandigitals here: http://www.iread.it/panmult.php
There are also several (12,289) pandigital numbers with more than one
pandigital
multiple. The 2 champions (which are pandigital times {2,4,5,7,8})
are 1,098,765,432 and 1,234,567,890.
giovanni resta
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