| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 510.1 | Some examples, not a proof ... | THEBUS::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri Jun 13 1986 16:10 | 38 |
| Well,
here are some info on the numbers 987654321... from calreal
________________________________________________________________________
CALREAL> is it a prime ( near prime of ( 9 ) );
.
.
.
CALREAL> is it a prime ( near prime of ( 98765432 ) );
Output:
-------
Nearest prime(s) to 9 is : 11, and : 7
9 is not a prime
Nearest prime(s) to 98 is : 97
98 is not a prime
Nearest prime(s) to 987 is : 991, and : 983
987 is not a prime
Nearest prime(s) to 9876 is : 9871
9876 is not a prime
Nearest prime(s) to 98765 is : 98773
98765 is not a prime
Nearest prime(s) to 987654 is : 987659
987654 is not a prime
Nearest prime(s) to 9876543 is : 9876553
9876543 is not a prime
Nearest prime(s) to 98765432 is : 98765431
98765432 is not a prime
________________________________________________________________________
Enjoy,
Kostas G.
|
| 510.2 | No too difficult | MODEL::YARBROUGH | | Fri Jun 13 1986 17:21 | 7 |
| 1) All the even G's are divisible by 2
2) All the G's ending in 5 are divisible by 5
3) All groups of 3 consecutive digits are divisible by 3, so the
G's ending in 7 and 1, which are position 3*n, are divisible by
3.
4) The 9's and 3's occur at position 3*n+1, so the G's ending with
3 and 9 are divisible by 3.
|
| 510.3 | alot easier than 123..., isn't it! | SIERRA::OSMAN | and silos to fill before I feep, and silos to fill before I feep | Fri Jun 13 1986 20:07 | 15 |
|
Right! Too bad we can't apply this sort of thing to problem
505.
Actually, I have some little "pamphlets" at home full of
math puzzles. I read the 987... one just a few days before
Mr. Gilbert posed the 123... one so I was wondering if
it was pure cowinkydinky or not.
Also, his other problem of 10^0 + 10^1 . . . was similar
to another I read in the pamphlet, so maybe we're looking
at the same source ?
/Eric
|
| 510.4 | | CLT::GILBERT | Juggler of Noterdom | Fri Jun 13 1986 21:52 | 2 |
| The problem just crossed my mind after doing a little work with note 501,
and 504 was related as a warm-up.
|
| 510.5 | Now that we know they are composite... | AKQJ10::YARBROUGH | Why is computing so labor intensive? | Thu Feb 25 1988 19:57 | 25 |
| Next (somewhat related) question:
Since all the permutations of the digits 123456789 are known to be
composite, (a) which has the largest number of distinct prime divisors? (b)
Which the smallest? (c) Which the minimum largest prime divisor [which may
be the same as the answer to (a)]?
A tentative answer to (a) follows the <ff>.
2*3^5*11*31*59*101 = 987561234 (six distinct divisors)
A very pretty answer to (b) appears on the next page:
There are twin solutions:
3^2*15873071 = 142857639
3^2*15873107 = 142857963
There are many others with two divisors. Is there a permutation that is the
product of two primes? No, because every perm. is divisible by 9.
A tentative answer to (c) follows the <FF>:
2^4*3^2*13*17*67^2 = 142857936 (minimax = 67)
|
| 510.6 | | CLT::GILBERT | Builder | Thu Feb 25 1988 21:52 | 24 |
| How many permutations have exactly n factors (not necessarily distinct)?
The following table shows this.
3 22366
4 69476
5 91799
6 76987
7 48781
8 26969
9 13655
10 6693
11 3168
12 1595
13 754
14 333
15 156
16 87
17 27
18 20
19 9
20 3
21 1
22 1
|
| 510.7 | Found Circa 1982 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Fri Feb 26 1988 13:44 | 6 |
| Re .5:
(c) 2^12 * 3^2 * 7^5 and 2^7 * 3^2 * 7^7.
-- edp
|