Math Fun

The dice game.

Each round of a dice game consists of two fair dice; the result of one throw is the product of the thrown numbers. A game consists of 5 rounds.

Bob throws in the second round by 5 more than in the first, in the third round by 6 less than in the second, in the fourth round by 11 more than in the third, and in the fifth round by 8 less than in the fourth.

How many points did he score in each of the 5 rounds?

The reputable house
[tab]2P
1G
1B

2P
1G
2B

2P
2G
2B[/tab]

Edit: I should post the process I guess

[tab]C > P > B > G > F

Because of the limit on the number of sisters a girl can have, from this I deducted that these have to be 2 parent families cuz there’s too many damn parents.

C = B + G
P = F2
G < F2

(B+G) > F*2 > B > G > F

I knew that F was going to be small so after this I tried out some values of F because I got lazy.[/tab]

That is false.

Please read the text one more time.

Dang, left out the “One family has more children than all the other families combined.” part.

Yes.

I feel like I must be missing something:[tab]3 families, each with 2 parents:
p = 2 2 2 = 6
g = 2 0 2 = 4
b = 1 1 3 = 5
c = 3 1 5 = 9[/tab]

[tab]1 3 = 4
2 7 = 9
3 4 = 7
4 1 = 5
5 1 = 6[/tab]

No. It doesn’t. You keep thinking that just because you have an operable algorithm, you have the only possible solution. For all of that type of problem, you MUST prove that your proposed solution is the only possible solution, else it isn’t a solution.

All of your deductions include an assumption that must be validated by being the only assumption possible in order to cause the puzzle to be solvable.

Proving that you have the one and only possible solution proves that the master is using it.

Definitely not true.

Well done, James. =D>

What about the solution process?

That is false. Please read the text one more time.

The reputable house

Duh myself
[tab]2P
1B

2P
2G
1B

2P
2G
3B[/tab]

Well done, Phoneutria. =D>

What about the solution process?

The dice game

[tab]a
b = a+5
c = b -6
d =c + 11
e = d - 8

c= (a+5) - 6
d = [(a+5) - 6] + 11
e = {[(a+5) - 6] + 11} - 8
max total=12

Round d can only be 11+1.

a=2
b= 7
c=1
d=12
e=4[/tab]

I just shuffled things around a little bit. I knew I couldn’t have been too far off because of the limit on the number of sisters.

ESP.

Damn … looking at my scratch work, it looks like I rewrote the whole puzzle … [tab]2 *5 =10
3 *5 =15
3 *3 =9
4 *5 =20
2 *6 =12[/tab]

That is false.

Please read the text one more time.

Well done, James. =D>

What about the solution process?

Hey, my job is to answer questions. Your job is to figure out how I got the answers.

[size=85]… just be sure to always disagree at least once so that I have incentive to go check my work …
[/size]

lol… the product, not the sum!
Facepalm…

Let me guess twice:

1. By reading, understanding, thinking, and calculating.
2. By finding the answers in the internet.

If I could look up the answer, I could look up the process for obtaining it.
I don’t go to that trouble for mere games.

1. I’ve presented a logical syllogism, using mathematical induction, which is a method of deductive mathematical logic.
2. As we’ve already discussed, once you have a syllogism, you don’t need to show that there are no other syllogisms (see e.g. the Pythagorean Theorem)
3. Even if it were the case that I needed to show there were no other solutions, for the MI problem I’ve provided a syllogism that shows that N islanders cannot learn their eye color before day N, so if there were another syllogism, it would produce the same result.

It is true for certain premises. If we assume X and ~X, we can conclude Y and ~Y. I’m saying that “this problem is not impossible” is an exotic premise that could similarly produce contradictory conclusions. I don’t think it does, but much of your argument seems to depend on it.

Right, but you’ll notice that I only said that the syllogism is “clearly valid”. The point being that a valid syllogism can be constructed using “this problem is not impossible” as a premise, and challenging you to show that the syllogism shouldn’t be treated like any other.

Do you agree that “colors don’t bear any logical relation to each other” is a true premise? Or do you have a similarly true premise that leads to a contradictory conclusion?

Only AFTER you skipped over substantiating your most essential premise (your presumption of colors) as well as your presumption that a faster method of color discovery could not be used.

Carl, that is stupid. Why do you keep saying that? Do you not read criticisms of your propositions? That is just a dumb thing to say and I have explained why. Once AGAIN:
[size=150]We are NOT talking about many proofs for the same outcome.[/size]

We are talking about the need to prove that there cannot be any other possible [size=150]outcome[/size], and thus no other possible algorithm/solution.

No, you have NOT. I showed you that everyone could easily leave after the second bell. You are simply not listening.

No, it isn’t.

It would be true if you made sense of it first, but as stated and knowing the context, I have to deny it. Colors have a natural relative ORDER (most often expressed as a frequency). An order isn’t technically “a logical relation”, but it is an association that can be used in a logic argument: “Green is between yellow and blue. Purple is between red and blue. Orange is between red and yellow.”
One of the premises is that “it is solvable”, so…

And also the presented colors CAN have a displayed order: