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[fractalman]

Log(both sides).

then you get log(x^logx)=1.
then, properties of logs let you say:
(Logx)(logx)=1.
log(x) must=+/-1.
then, [x=10, x=1/10. ]

The rest of you missed a possible solution. .


As for God's minimum wattage:
"And God said 'let there be light' and there was light. ".  Have fun.  (max time=1 day. minimum lighting amount=visible universe.)

[MagmaMcFry]

then, [x=10, x=1/10. ]

The rest of you missed a possible solution. .

No we didn't.

[Skyrunner]

God's minimum wattage:
"And God said 'let there be light' and there was light. ".  Have fun.  (max time=1 day. minimum lighting amount=visible universe.)

Go ahead, find the volume of the entire universe  It's error:undefined here.

Besides, it's silly abuse of Biblical phrases and math anyways.

[MonkeyHead]

God's minimum wattage:
"And God said 'let there be light' and there was light. ".  Have fun.  (max time=1 day. minimum lighting amount=visible universe.)

Go ahead, find the volume of the entire universe  It's error:undefined here.

Besides, it's silly abuse of Biblical phrases and math anyways.

Meh, I think we would only need the volume of the observable universe (which of course is trivial...) - if it is unobservable it clearly has no "power" that we can be influenced by.

[Pnx]

It's late... I'm tired, I've tried to figure out how to do this one on my own and it's been a bit frustrating, if someone could give a good explanation of what I'm doing and why I'm doing it that I can read when I get up in the morning it would be much appreciated:

Use the given vectors to find the value of the indicated scalar.
u=-i+5j , v=3i-5j , w=-3i+j
Find u·w + v·w

[Karlito]

It's just dot products. Do you know the dot product? If u = ai + bj and v = ci +dj then the dot product
u·v = ac+bd.

So in your case u·w + v·w = -1*-3 + 5*1 + 3*-3 + -5*1 = -6

[Jim Groovester]

The dot product is the sum of coordinate-wise multiplication between two vectors. i and j are the typical names for the unit vectors in R² and R³, in physics anyway, with i = (1,0) and j = (0,1).

If you're taking a linear algebra class instead of a physics class, you'd represent u, v, and w, in the following ways:

u = (-1,5)
v = (3,-5)
w = (-3,1)

And therefore,

u·w = -1 * -3 +  5 * 1 = 8
v·w = 3 * -3 + -5 * 1 = -14

u·w + v·w = 8 + -14 = -6

FAKEEDIT: Goddammit, ninja'd.

[Vector]

1/6 + 5/6*1/6 + 5/6*5/6*5/6*1/6 + . . .

= 1/6 + Sigma 1/6[(5/6)]²ⁿ⁺¹ over n = 0 through infinity

[rhesusmacabre]

Well, here's a problem that I did manage to solve, but I'm not sure whether it is the best approach...

Problem: Find the sum of all three-digit integers not divisible by 3.

Do you mean positive integers or all integers? The second case would cancel to zero, so assuming the first:

The integers can be paired up in pairs of sums like so:

100+998=1098, 101+997=1098,
103+995=1098, 104+994=1098,
 ∶
547+551=1098, 548+550=1098.

Now 100=(3*33)+1, 998=1098-((3*33)+1), 101=(3*33)+2, 997=1098-((3*33)+2),
and  547=(3*182)+1, 551=1098-((3*182)+1), 548=(3*182)+2, 550=1098-((3*182)+2).


If a_n=(3n+1)+(1098-(3n+1))+(3n+2)+(1098-(3n+2)) = 2*1098 = 2196,
then Σ_n=33¹⁸² a_n = (182-32)*2196 = 329400.

[Soadreqm]

Problem: Find the sum of all three-digit integers not divisible by 3.

This is the kind of problem that I would solve numerically, by just telling a computer to sum all the integers together. In fact, I just did.

Code: [Select]

#Antti Virkkunen
#2013-05-05
#This is Python by the way
s = 0
i = 0
for i in range(100, 1000):
    if (i%3 != 0):
        s = s+i
print(s)I got 329400.

[MagmaMcFry]

Problem: Find the sum of all three-digit integers not divisible by 3.

That's just the sum of all integers from 100 to 999, minus thrice the sum of all integers from 34 to 333. Gauss's formula says this is equal to (499500-4950)-3*(55611-561) = 329400.

Then there's a second problem that has me rather stumped:

Two players, A and B, are playing a game. One player rolls a fair six-sided die, trying to get a six. Then the second player does. They alternate until one gets a six. Find the probability that player A will get the first six, assuming he starts.

Let's say that a is the probability of A winning, and b is the probability of B winning.

Obviously, a + b = 1.
Now if A doesn't roll a 6, then B will have the same chance of winning as A had at the beginning of the game:
b = a * 5/6.
Substitution gives a + a * 5/6 = 1, therefore a = 6/11. Simple as that.

[Skyrunner]

Spoiler: Magma's answer (click to show/hide)

Then there's a second problem that has me rather stumped:

Two players, A and B, are playing a game. One player rolls a fair six-sided die, trying to get a six. Then the second player does. They alternate until one gets a six. Find the probability that player A will get the first six, assuming he starts.

Let's say that a is the probability of A winning, and b is the probability of B winning.

Obviously, a + b = 1.
Now if A doesn't roll a 6, then B will have the same chance of winning as A had at the beginning of the game:
b = a * 5/6.
Substitution gives a + a * 5/6 = 1, therefore a = 6/11. Simple as that.

Spoiler: Vector's answer (click to show/hide)

1/6 + 5/6*1/6 + 5/6*5/6*5/6*1/6 + . . .

= 1/6 + Sigma 1/6[(5/6)]²ⁿ⁺¹ over n = 0 through infinity

I'm not sure which one is correct, 6/11(~0.54) or 0.62. But I tend to think that it's Vector's, 0.62, because I believe Magma only iterated once.

[ed boy]

1/6 + 5/6*1/6 + 5/6*5/6*5/6*1/6 + . . .

= 1/6 + Sigma 1/6[(5/6)]²ⁿ⁺¹ over n = 0 through infinity

I'm not sure which one is correct, 6/11(~0.54) or 0.62. But I tend to think that it's Vector's, 0.62, because I believe Magma only iterated once.

Not quite, it's 1/6+5/6*5/6*1/6+5/6*5/6*5/6*5/6*1/6 + ... =Sigma 1/6[(5/6)]²ⁿ for n from 0 to infinity, which indeed is equal to 5/6.

[ZetaX]

The explaination why both are correct is that Vector just did it the "obvious" way (sum up all possibilities with the corresponding probabilities), while Magma made use of a symmetry to make the argument nicer.

[Skyrunner]

I think Magma is wrong.
His answer only had then rolling it once each, I think, while the problem said they kept rolling until one side won.