| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 209.1 | | FUTBAL::BUDNIK | | Wed Jan 16 1985 09:45 | 10 |
| Assuming that you want to eliminate the trivial solution(s) with a
single digit the answer is:
10a + b = 8b + a
9a = 7b
a/b = 7/9
smallest integer values for a & b are 7 & 9 so
decimal 79 = octal 97
|
| 209.2 | | AURORA::HALLYB | | Wed Jan 16 1985 10:33 | 1 |
| Octal 97?
|
| 209.3 | | FUTBAL::BUDNIK | | Wed Jan 16 1985 11:25 | 1 |
| Oops, sorry about that. Back to the drawing board!
|
| 209.4 | | METOO::YARBROUGH | | Wed Jan 16 1985 13:10 | 1 |
| Hmm. If you allow leading 0's the smallest non-trivial solution is 03160.
|
| 209.5 | | FUTBAL::BUDNIK | | Wed Jan 16 1985 13:09 | 11 |
|
decimal 1527465 = octal 5647251
Being a programmer rather than a mathematician I gave up trying
to find the answer myself and just asked my friendly computer.
(I find that is usually more reliable). However, I would still
be interested in seeing an "elegant" mathematical solution.
|
| 209.6 | | HARE::STAN | | Wed Jan 16 1985 13:41 | 6 |
| 1527465 is indeed correct and is the smallest solution.
Only one of many people solved this problem in the Journal of
Recreational Mathematics, and he, apparently, used a computer too.
Quick work!
|
| 209.7 | | TAV02::NITSAN | | Thu Jan 17 1985 00:35 | 6 |
| d d-1
If n has d digits (base 8 and base 10) then 8 > n >= 10 ,
so d*log(8) > (d-1)*log(10) , so d < log(10)/(log(10)-log(8)) ,
so d<=10...
|
| 209.8 | | SPRITE::OSMAN | | Thu Jan 17 1985 18:22 | 9 |
| Is there a related problem in NON-integers here ? For instance, is
it interesting to look for
0 . d1 d2 d3 . . . dn (10) = 0 . dn . . . d3 d2 d1(8) ?
Or perhaps
d1 d2 d3 . . . dn . e1 e2 e3 . . . em (10) =
em . . . e3 e2 e1 . dn . . . d3 d2 d1 (8) ?
|
| 209.9 | | R2ME2::GILBERT | | Fri Jan 18 1985 19:05 | 8 |
| A couple other possible problems.
Since the number of digits is bounded (see a previous response), there
shouldn't be very many solutions. What are they?
The same problem can be posed for different pairs of bases. For example,
for bases 3 and 5, we have 21 (base 3) = 12 (base 5). Are there any pairs
of bases (not greater than base 16) that have no 'reversible' number?
|
| 209.10 | | R2ME2::GILBERT | | Fri Jan 18 1985 23:22 | 20 |
| Here are some known solutions for other pairs of bases, up through base 16.
An asterisk means there's no solution. A question mark means that one is
still 'open'. The open ones are: (6,7), (8,9), (10,11), (10,12), (12,14),
and (14,15).
3 4 5 6 7 8 9 10 11 12 13 14 15
3 X
4 112 X
5 12 * X
6 * * 223 X
7 * 12 23 ? X
8 * * 133 * 334 X
9 * * 12 255 34 ? X
10 * 13 * * 23 1527465 445 X
11 * * * 12 35 122 45 ? X
12 * * * * * 337 14124 ? 556 X
13 * * 13 115 12 10107 23 34 56 66BB8 X
14 * * * * * * 225 438 355 ? 667 X
15 * * * * 125 12 47 * 57 588 67 ? X
16 * * * 13 25 * 1075 35 23 538 45 123D1 778
|
| 209.11 | | TURTLE::GILBERT | | Sun Jan 20 1985 02:40 | 3 |
| Here are two more: (10,12) 132513, and (14,15) 1528C1.
The following are still open: (6,7), (8,9), (10,11), and (12,14).
|
| 209.12 | | AURORA::HALLYB | | Mon Jan 21 1985 18:27 | 1 |
| Then there's the equivalent problem for negative number bases ...
|
| 209.13 | | GOLLY::BUDNIK | | Tue Jan 22 1985 09:12 | 3 |
| here's one more: (10,11) 454003312
that leaves: (6,7), (8,9), (12,14)
|
| 209.14 | | TURTLE::GILBERT | | Tue Jan 22 1985 23:53 | 3 |
| One more: (6,7) = 10541322115
Leaving (8,9) and (12,14).
|