| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1643.1 | Any cute solutions? | VMSDEV::HALLYB | Fish have no concept of fire. | Wed Jul 15 1992 12:05 | 10 |
| This would seem to be nothing more than asking if the multiplication
table supplied is associative. It is:
axy = xay = xya = a for any x,y so anything with an a is OK
bxy = xby = xyb = xy for any x,y so anything with a b is OK
There must be somtehing clever about the others, but I just verified
they were associative by exhaustive search.
John
|
| 1643.2 | | ZFC::deramo | Dan D'Eramo | Wed Jul 15 1992 13:42 | 7 |
| It didn't say that the letters on the board were in any order.
You just pick two of them, any two, not necessarily adjacent,
to be combined. So having the operation be commutative (it is
easy to see that it is) would help. As would having an 'a' on
the board. :-)
Dan
|
| 1643.3 | | BEING::EDP | Always mount a scratch monkey. | Wed Jul 15 1992 13:47 | 5 |
| Hmm, I didn't think it was associative, so I assumed there was
something a bit trickier. Well, I'll enter another problem.
-- edp
|
| 1643.4 | | BEING::EDP | Always mount a scratch monkey. | Wed Jul 15 1992 13:49 | 6 |
| 1. Find all real solutions (x,y,z) satisfying the simultaneous
equations
(x+2y)(x-2z) = 24,
(y+2x)(y-2z) = -24, and
(z-2x)(z-2y) = -11.
|
| 1643.5 | | KOBAL::GILBERT | Ownership Obligates | Wed Jul 15 1992 14:25 | 9 |
| re .2:
Commutativity doesn't suffice. To wit, consider:
a b c
a a c b
b c b c
c b c a
Then (aa)b = ab = c, while a(ab) = ac = b.
|
| 1643.6 | | ZFC::deramo | Dan D'Eramo | Wed Jul 15 1992 19:42 | 11 |
| > Commutativity doesn't suffice.
Of course not. .1 said
> This would seem to be nothing more than asking if the multiplication
> table supplied is associative. It is:
In .2 I added commutativity as another condition, in addition
to associativity.
Dan
|
| 1643.7 | simple solution to .0 | DESIR::BUCHANAN | | Thu Jul 16 1992 07:45 | 7 |
| The point of the puzzle in .0 is that {a,b,c,d,e} maps onto the
integers mod 5 under multiplication. To be specific:
a->0, b->1, c->4, d->2, e->3
Since integer multiplication is commutative and associative (see D'Eramo
1562.2 for a sketch of the proof :-) ) this model proves the result for us.
Andrew
|
| 1643.8 | trivial with MAPLE | DESIR::BUCHANAN | | Mon Jul 20 1992 10:59 | 31 |
| Exploration showed that there *were* solutions for y=0, so I tried a
frontal assault using MAPLE.
> f1 := (x+2*y)*(x-2*z)-24;
f1 := (x + 2 y) (x - 2 z) - 24
> f2 := (y+2*x)*(y-2*z)+24;
f2 := (y + 2 x) (y - 2 z) + 24
> f3 := (z-2*x)*(z-2*y)+11;
f3 := (z - 2 x) (z - 2 y) + 11
> solve({f1=0,f2=0,f3=0},{x,y,z});
{z = 1, y = 0, x = 6},
{z = -1, y = 0, x = -6},
{x = -8/3, z = -13/3, y = 10/3},
{y = -10/3, x = 8/3, z = 13/3},
2 4
11173 117 3 9811 %1 + 351 %1 - 14982
{y = ----- %1 + --- %1 , x = 1/2 --------------------------, z = %1}
2724 908 2
%1 (9811 + 351 %1 )
4 2
%1 := RootOf(729 _Z + 22734 _Z + 51529)
But this quartic obviously has no real roots, so we have just the
four solutions quoted before.
Andrew.
|
| 1643.9 | | RUSURE::EDP | Always mount a scratch monkey. | Thu Jul 22 1993 14:29 | 14 |
| The answers here are confirmed by the published answers. For problem
1, the published answer writes x+y-z=a, y+z=b, and z+x=c, from which
the equations reduce to:
a^2-b^2=24, a^2-c^2=-24, and a^2-(b-c)^2=-11.
Eliminating a^2 gives c^2-b^2=48 and c^2-(b-c)^2=13, from which
c=(b^2+13)/2b and 3b^4+166b^2-169 = (b^2-1)(3b^2+169)=0. Thus b=1 or
b=-1. When b=1, c=7 and a=5 or a=-5. When b=-1, c=-7 and a=5 or a=-5.
From the first substitution one finds four solutions for x, y, and z,
as listed in .8.
-- edp
|