| 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 |
Proposed by Tim Cross, Wolverly High School, Kidderminster, U.K.
(a) Find all positive integers p <= q <= r satisfying the equation
p + q + r + pq + qr + rp = pqr + 1.
(b) For each such solution (p, q, r), evaluate
atan(1/p) + atan(1/q) + atan(1/r).
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1902.1 | SSAG::LARY | Laughter & hope & a sock in the eye | Thu Oct 06 1994 13:16 | 17 | |
I gotta run to a meeting, but it looks like the equality can be rewritten as (p+1)(q+1)(r+1) = 2(pqr+1) dividing by pqr and noting that p <= q <= r, we get (1 + 1/p)^3 >= (1 + 1/p)(1 + 1/q)(1 + 1/r) = 2 + 2/pqr > 2 which means p must be 1, 2, or 3. That gives three subcases for q and r, which can be similiarly limited. The p=1 case is quickly found to have no solutions. I ran out of time to try p=2 and p=3, but a quick computer search indicates that the only soluions for q <= r <= 1000 are: p=2, q=4, r=13 p=2, q=5, r=8 p=3, q=3, r=7 | |||||
| 1902.2 | IOSG::TEFNUT::carlin | Dick Carlin IOSG Reading | Thu Oct 06 1994 14:20 | 33 | |
p = 2 means that
1+3q+3r = qr
so q = (3r+1)/(r-3)
= 3 + 10/(r-3)
so r = 4,5,8 or 13
q = 13,8,5 or 4 two distinct solutions.
similarly one distinct solution for p=3, so you found them all.
(The second half)
tan(a+b+c) = tana+tanb+tanc - tanatanbtanc
-----------------------------
1 - (tanatanb+tanbtanc+tactana)
so if p=1/tana etc we get
tan(a+b+c) = pq+qr+rp - 1
------------
pqr - (p+q+r)
= 1
so atan(1/p)+atan(1/q)+atan(1/r) = a+b+c = pi/4 +/- n*pi
and, by inspection :-) for the p,q,r values found, it is pi/4
whereas (by using p=tana etc) it looks as if
atan(p)+ata(q)+atan(r) might be 5pi/4
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| 1902.3 | RUSURE::EDP | Always mount a scratch monkey. | Fri Aug 11 1995 14:09 | 8 | |
The published solution agrees with the previous responses.
-- edp
Public key fingerprint: 8e ad 63 61 ba 0c 26 86 32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.
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