harey

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Point  1 light
Can you prove it?
I hesitate between 1 and 2...

Well, I am just not very sure concerning the assumption in b). I do not consider whether there is a solution of another kind  when solving a problem, there is a little bit of intuition on the beginning. This does not make me doubt about the solution if I find one, except in the case I would claim it is unique.
As for the remainder, there are so few possible solutions (we look for integers) that you can try them all. Otherwise, you could solve it as a system of multiple equations with multiple variables, but I think it is too much hassle here.
I am awaiting the official solution!!!

a) 1 on end means (...+1)(....1)
b) as there is no term with power 3, I made the bet there only are 2 termsc) multiplied (ax+by+1)(cx+dy1)
d) ac=2; (ca)=1; I would find very tasteful a=2; c=1;
e) bd=2; (db)=3; b=1 and d=2 ???
f) (ad+bc)=(2*21)=3: It works!!!
FInal form:(2xy+1)(x+2y1)

Bob can refine this strategy. Suppose:
a) the number of apples is limited to N (a very large one) and
b) Alice places places a random number of apples in each room
 if the number of apples in the first room is "small", Bob should refuse it
 if the number of apples in the first room ">N" and the number of apples in the second room is "just a little bit smaller than the number of apples in the first room", Bob should accept it
If we lift a), it will still apply.
Alice can counter this strategy by placing respectively K, K+1, K+2 apples. This brings us back to the strategy of kman.
The mathematical formulation is left to the reader as exercise
However, if Bob always accepts the first room and decides to accept or reject a second room based on whether the number of apples in the second room is larger or smaller compared to the first room then his chances of winning become 50%.

Call the ball with the lowest mass L.
Call the ball with the highest mass H.
D=mass(H)mass(L)
D=sum of masses of greens going from L to H clockwise
D=sum of masses of greens going from L to H counterclockwise

I get the same result as bonanova and Yoruichisan.
Wanting to avoid algebraic errors, I asked Wolfram Alpha:

Let's call Wilhelm the kid
 in the same column as Pete (Pete is older than Wilhelm)
AND
 in the same row as Jane (Wilhelm is older than Jane) 1

Draw a circle. Inscribe a square ABCD, at best AC being horizontal. Move the points A and C some degrees towards D, delete the arc AC. Move D some 4 units from A. You get a kind of hot air baloon or an ice cream cone.
#define RAC = Remaining Arc of the Circle
 3 points on the RAC: 4th point must be on the circle, at least 90 degrees missing;
 3 points on the cone: 4th point cannot be on the cone (lines not parallel) nor on the RAC;
 2 points on the cone and 2 points on the RAC: draw two parallel lines, resulting segments can never have the same length
It is hard to inscribe a square into a regular pentagon. I conjecture that many polygons will fit.

1) number the computers 1 to N
2) test by the computer i=max(1,last_tested) the computer (i+1)
3) if the answer is bad, discard both (and renumber)
4) go to 2 unless one of these cases occurs:
A) You discarded 2*k computers => as you always discarded a good one and a bad one, the remaining [one is /are] good.
B) You got k answers "GOOD", There cannot be a bad computer after a good one; there might well be a series a bad ones (even 0) followed by good ones (even 0). As there are at most k bad ones, (k+1)th is good.
C) A mix of A and B... (Easy to deduct, hard to describe w/o loosing simplicity. Do not forget to adapt k.))
In any case, k tests are enough.

To be clear the sexual activity will occur in a round robin sense
What does it mean?
I have never been to an orgy

I do not really fancy this kind of problem. The solution usually is "What would answer the god B if the god A asked him what would answer the god C if..."
Did you think about something like that?

Assuming that Santa must always get progressively closer to the south pole. What is the slowest (longest duration of travel) he can travel and remain forever at night? So far both you (bonanova and TSLF) claim it is 1 year. I am asking if this is correct.
I am assuming we are looking for the smallest CONSTANT speed.
If we want to maximize the travel time (at varying speed), he can travel almost forever: he follows the shadow line moving imperceptibly to the south.

I did not really get why the 12 hours...
 by symmetry, we can calculate only the trajectory to the Equator
 if Santa moves faster then the speed of the shadow line, he should move West as much as possible (I can develop if necessary)
 his move to the West is limited by the shadow line
( when his speed equals the speed of the shadow line, he should make some turns around the Earth until the South pole gets into the shadow  solves Bonanova's remark, but we can omit this in the calculation)
 when his speed is smaller than the speed of the shadow line, he should move plain South
When he moves to the West, he trades somehow time against distance, but I cannot find the equivalence.

The fastest 5 runners can be determined in 7 heats.
....
I'll wait for discussion before posting the 7heat solution
Please post, I think that you cannot get more than 3:
Make them run 5 by 5, attribute a letter to each race and number them in the order of arrival. You get:
A1 A2 A3 A4 A5
B1 B2 B3 B4 B5
...
E1 E2 E3 E4 E5
6th heat: Make run A1, B1, C1, D1, E1. If they do not arrive in that order, reletter them so A1 is faster then B1 who is faster than C1...
Gold medal: A1
Other medals: make run (7th heat): A2, A3, B1, B2, C1.
If they arrive in this order, the 4th fastest can be A4, D1...

No.
Behind the first bullet, there always is a combination of bullets which contains one that will reach the first one.

N1 (except for N=3 or 4 where you can do better).
Let's put M=(N1)/2.
If we ask all i to test all i+1, the most unlucky case occurs when all answers are GOOD. Consider that all bad computers are followed by all good ones. We only can be sure the last one is good  there is no bad computer after a good one.
If the answer is BAD:
 discard both (you never discard two good computers)
 substract 1 from i (if i=0, take anyone)
 renumber...
Now, each computer said the next one is GOOD, so we are in the case (B)(B)(B)..GGGG.
In the worst case:
a) we always discarded a good computer and a bad one
b) both were tested
c) there were M bad computers and therefore M discardments
d) b+c imply 2*M tests.
If a bad computer says BAD, the above still applies.
I just hope I did not forget something...
@Rainman
Suppose:
a) BGBXX
b) BBGXXc) GGBXX
The results of the tests can be GOOD and BAD in each case. 3 and 4 are special cases, so you cannot use them for induction.

near N*2.
Test A by B and B by A. One of the following happens:
a) bad/bad: at least one computer is bad, put them both aside;
b) good/bad (same as bad/good): idem;
c) good/good: both are good or both are bad; put one aside and keep one.
You never put aside 2 good computers, so you still have more good computers that bad computers. As the worst case is c), we put aside N/2 computers with N tests. (If the number is odd, keep one it for the next round.)
Proceed the same way with the computers you kept.
When you have only 3 or 4 computers, one (reciprocal) test is enough.

We already had this problem, something like drawing two different fruits from a handbag. I just cannot locate it.
http://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem

If our hardware uses fewer clock cycles to perform three additions than a single multiplication, we may well gain overall processing speed by using Eq. (24) and Eq. (25) instead of Eq. (12) for complex multiplication
If it worked, it would be great for fractal pictures. The bad news is that multiplications are in the processor heavily optimized, a fair guess 45 additions. Googling:
The latency is 1 cycle for an integer addition and 3 cycles for an integer multiplication. You can find the latencies and thoughput in Appendix C of the "Intel 64 and IA32 Architectures Optimization Reference Manual", which is located on http://www.intel.com/products/processor/manuals/.
The big question is how the 3 temporary variables were treated by the compiler. If we have to access the RAM...

@bonanova  sorry, but I believe you have a sign error with the factor bd:
I = (a+b)(c+d)  R =ac+bc+ad+bd  (ac  bd) ... = ... +2*bd ....

@bmad
As nobody found the solution, can you post it?

Yes, I should have started at 0. (a,b)=(0,6) > (a+b)=6

1) sum of ages=boys+girls+teacher=b*g+g*b+42=2*b*g+42
2) sum of ages=average_age*number_of_persons=(b+g)*(b+g+1)
1=2: 2*b*g+42=(b+g)*(b+g+1)
I fed it into Wolfram Alpha and got no positive integer (number of persons) solution.
A small program to be sure:
for b in range(1,99):
for g in range(1,99):
sum1=2*b*g+42
sum2=(b+g)*(b+g+1)
if(sum1==sum2):
print(b,g,sum1)
(b,g)=(3,5) => b+g=8
I would be interested if someone finds the solution with Wolfram Alpha.
Sorry, unable to hide. 
... B.
If we play the diagonal, I win. The losing places are A1, C3, D5, E7.
If A plays right (up), I play right (up), too. Sooner or later these moves will be compensated by my opponents up (right) and mine up (right).
Maybe factoring a polynomial completely
in New Logic/Math Puzzles
Posted
This is just a typo, sorry for that:
for a in [2,1,1,2]:
c=int(2/a)
for b in [2,1,1,2]:
d=int(2/b)
if(db==3):
if(a*d+b*c==3):
print("a=",a,"c=",c,"b=",b,"d=",d,"ad+bc=",a*d+b*c,"db=",db)
print(a*c,"x2+",a*d+b*c,"xy+",b*d,"y2+",ca,"x+",db,"y1")
The problem does not have to be solved by a system of equations,
As no one else posts and I have no other idea how to solve it, can you post the solution?