| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
(1) Let S = 1 + 1/2 + 1/3 + ... + 1/p, where p is a prime greater
than [or equal to] 5. Show that S can be written in the form
r/(ps), where r and s are positive integers, and that
r-s = 0 (mod p^3).
(2) Prove Wolstenholme's theorem: if p is a prime greater than
[or equal to] 5, then
( 2p-1 )
( p ) = 1 (mod p^3).
(4) Find the limit points of the following sequences:
(a) {n*sin(2*pi*n!*e)}
(b) {n*sin(2*pi*n!/e)}
(5) Find the largest even number that can't be written as the sum
of two odd composite numbers.
Chris Long
Rutgers University
3. An old but nice problem. Find all real solutions to the following:
x x x
(x+1) + (x+2) = (x+3) .
Ambati Krishna
Electrical & Computer Engineering and
Mathematical Sciences Departments
The Johns Hopkins University
Baltimore, MD 21218
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 787.1 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Mon Nov 09 1987 18:43 | 6 | |
Can somebody figure out the correction for the first problem? As
stated, S is obviously positive and rational, so it is u/v for some
positive integers u and v. Then let r = p^4*u and s = p^3*v.
-- edp
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| 787.2 | Probably meant (r,ps)=1 | ZFC::DERAMO | Daniel V. D'Eramo | Mon Nov 09 1987 21:38 | 3 |
Without having done any work on the problem, I would assume that
the representation r/(ps) was supposed to be reduced so that r and
ps were relatively prime.
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| 787.3 | Not a proof, just an answer | ZFC::DERAMO | Daniel V. D'Eramo | Mon Nov 09 1987 22:35 | 26 |
>> 3. An old but nice problem. Find all real solutions to the following:
>> x x x
>> (x+1) + (x+2) = (x+3) .
What follows is not a proof, merely what I found by a quick examination
of a few trial cases.
It seems like the only solution over the reals is x = 2; substituting
2 for x gives 3^2 + 4^2 = 5^2 -- gee, that looks familiar!
If x = -1, -2, or -3, then zero is being raised to a negative power,
and so I am ruling those x's out. If x < -1 and is not an integer,
then (x + 1)^x is a negative number raised to a fractional power,
so out they go, too. Then I tried -4, -5, -6, ... and the values
of
(x+1)^x + (x+2)^x - (x+3)^x
did not become zero. For x > -1 there were no domain problems,
and again by examining values I found the root x = 2 and it appeared
that there were no other roots.
I suppose a real proof would involve checking the derivative to
verify that the growth trends guarantee no other roots.
Dan
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| 787.4 | CLT::GILBERT | Builder | Tue Nov 10 1987 14:27 | 17 | |
| 787.5 | CLT::GILBERT | Builder | Thu Nov 12 1987 18:25 | 22 | |
(5) Find the largest even number that can't be written as the sum of two odd composite numbers. Solution follows. Let the even number be 2*m. Let x = (19*m + 15) mod 30. Then m+x mod 5 = (20*m + 15) mod 5 = 0, and m-x mod 3 = (15 - 18*m) mod 3 = 0, and m+x mod 2 = (20*m + 15) mod 2 = 1, and m-x mod 2 = (15 - 18*m) mod 2 = 1. Thus, m-x is an odd multiple of 3, and m+x is an odd multiple of 5, and (m-x) + (m-x) = 2*m. If m-x is greater than 3, then m-x is an odd composite number (it has another factor besides the 3), and if m+x is greater than 5, then it is also an odd composite number. This construction suffices for m > max(3+x, 5-x) >= max(31, 5) = 31. For smaller m, we find the decompositions: 62=27+35, 60=9+51, 58=9+49, 56=21+25, 54=9+45, 52=25+27, 50=15+35, 48=9+39, 46=21+25, 44=9+35, 42=9+33, and 40=15+25. But there is no such decomposition for 38 -- the answer -- each of the possible decompositions into two odd summands include a prime, viz: 38 = 19+19 = 17+21 = 15+23 = 13+25 = 11+27 = 9+29 = 7+31 = 5+33 = 3+35 = 1+37. | |||||