/sci/ - Science & Math

>>4380891
A^n always has odd along the diagonal, so it's always divisible by 2.
A^(2^n) - I has GCD of the GCD of A^(2^(n-1)) - I times the GCD of A^(2^(n-1)) + I - just factor out the GCDs and multiply, then multiply it back in, for proof.
But GCD A^(2^(n-1)) + I is divisible by 2, and GCD of A^(2^(n-1)) is divisible by 2^(n-1) by inductive hypothesis which I didn't actually state before, so GCD A^(2^n) is divisible by 2^n

The answer is 14.

The pattern tricks you into thinking 112.

>>4377626
In light of I'm a dumbass, I retract my statement.

>>4377397
Easy as pie.
First, we define distance from a vertex to be the min edges traversed.
Then starting at corner s, there are n vertices at distance 1, and there are n(n-1)/2 vertices at distance 2, and so on. In fact, it's pretty straightforward but a little mind-boggling that the number of resistors from a vertex at distance k to vertices at distance k-1 is k. You can see this empirically on the cube, and I'm not good with actually arguing my way around polytopes in some higher dimension, but if I don't fall asleep I'll come back and argue it as a lemma.

By symmetry, the points at distance 1 are indistinguishable, so they have equipotential.

So now we know that the resistance from level 0 to a vertex at level 1 without considering the other resistors is 1, the resistance from level 1 to level 2 is 1/2 because there are two incoming resistors HOLY SHIT IT'S THE HARMONIC SERIES, ISN'T IT.

>>4376665
Read: I worked NEAR people who programmed, and I'm therefore an expert.

>>4376647
Yup, that's basically 80% of the thread, with the other 20% being some idiot going "No, there's a difference between having learned it and having that knowledge just _programmed in_."

Fuck, I need a hug.

>>4376489
Oh, one more thing, before everyone else in the thread has harped on you for it - a definition can be wrong, if it is inconsistent with the axioms. You can't define f(x) to be the set of all numbers not in f(x), because it doesn't make any damn sense.

Likewise, you can't define one program to be a set of responses and the other to be something that organically acquired knowledge because any knowledge that could be acquired could be hard-coded in, and the hard-coded knowledge could be indistinguishable from any other sort of knowledge.

There is literally no way to distinguish between these two concepts as they've been described in this thread.

There

>>4376506
But you can't have a process outside of programming take place in a program, or even in a brain. A brain works a certain way, no more and no less. If there's something beyond programming, it's an intangible, a soul, and I reject that as a way to meaningfully distinguish between anything because I've never seen a soul and I don't know of anyone who has come up with an empirical test for a soul.

I think you should try saying what you wanted to say, but again and much more carefully.

>>4376413
If you mean hard-programmed response, where the machine can take the input and produce output then return to exactly the same state, suppose there is such a thing as hard AI. Freeze the state of a hard AI, and have it, as a program with a particular state, run on the Chinese inputs, then flush state and return to that original state. Now it's a weak AI. It doesn't understand, because it's following a set program. It doesn't learn, because there's no change in state that could contain any learned knowledge. It's reduced to weak AI.

Do the same thing with a brain. But the brain, by supposition, understands.

Now, go back to the strong AI. Hard-code every bit of the brain's functionality, but return to state every time. It's even a deterministic program. Perhaps you allow the state to persist. It's not a hard AI, because it's just following a set of rules to generate Chinese characters and then changing state.

None of these things are at all different, and yet one's a real brain, the golden standard of understanding, one's a hard AI, and one's a hard-coded set of responses and state changes.

I meant to say "distinguishable"; a machine that can't reason at all is distinguishable from a person.

>>4376313
No, there's no such distinction being made. If a person can make inferences, then the machine, given the same input, would have to be able to make inferences. Otherwise it's indistinguishable from a person.

This argument would fall apart if it were used to prove something rigorous in complexity theory. Just because something can be done in more than one way doesn't mean that all ways of doing it are wrong.

Say the human way of understanding Chinese involves subconsciously turning Chinese words into concepts, memories, and patterns of thought. Suppose a human can consciously compute the patterns of thought that a human goes through to understand Chinese. Suppose I do that for a Chinese speaker's brain. I don't understand Chinese; therefore, neither does the Chinese person.

The argument is flawed.

>>4374805
Well, prime numbers are the generators of the prime ideals of the integers, and prime number theory is extended to prime ideal analysis, where prime ideals have special properties in the category of ideals.

>>4371667
And whoever said it was wrong, this confirms the results above,
>>4370624
>>4370042

>>4371679
And by "one" I mean "none."

Sorry, guys, I wasn't in this thread until 5 minutes ago.

>>4371667
By "a lot" I mean "one"

If <span class="math">|n - m^2| \leq 250[/spoiler], then the range is 501 numbers; the squares lie between some number n and 500 more than that number. Because these numbers correspond one-to-one with the original n, I'll now only refer to the new number as n, and use the condition <span class="math">n \leq m^2 \leq n + 500[/spoiler]
So we have a number of conditions on the least a such that m = a solves this.

<span class="math">a^2 - 2 a + 1 < n[/spoiler]
<span class="math">a^2 \geq n[/spoiler]
<span class="math">a^2 + 28 a + 196 \leq n + 500[/spoiler]
<span class="math">a^2 + 30 a + 225 > n + 500[/spoiler]

By flipping the second condition and adding it to the third condition, we get
<span class="math">28 a + 196 \leq 500[/spoiler]
<span class="math">28 a \leq 304[/spoiler]
<span class="math">a \leq 10[/spoiler]

By flipping the last condition and adding it to the first condition, we get
<span class="math">275 < 32 a[/spoiler]
<span class="math">8.6 < a [/spoiler]

So a = 9, 10

If a is 10, then n is between 82 and 100, because <span class="math">9^2 = 81[/spoiler], and because <span class="math">24^2 = 576[/spoiler], <span class="math">25^2 = 625[/spoiler], there's no danger of running into 25. This gives us 19 values of n.

If a is 9, then n is 65 to 75, because <span class="math">24^2 = 576 > n + 500[/spoiler], <span class="math">8^2 = 64 < n[/spoiler]. This gives us 11 values for n.

All in all, there are 30 values. To get the original values of n, add 250, to get 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350.

Remember, if there are 15 squares, there are 14 more than whichever is your reference. A lot of these analyses seem to miss that.

Sorry I couldn't post yesterday.

If <span class="math">|n - m^2| \leq 250[/spoiler], then the range is 501 numbers; the squares lie between some number n and 500 more than that number. Because these numbers correspond one-to-one with the original n, I'll now only refer to the new number as n, and use the condition <span class="math">n \leq m^2 \leq n + 500[\math]
So we have a number of conditions on the least a such that m = a solves this.

<span class="math">a^2 - 2 a + 1 < n[/spoiler]
<span class="math">a^2 \geq n[/spoiler]
<span class="math">a^2 + 28 a + 196 \leq n + 500[/spoiler]
<span class="math">a^2 + 30 a + 225 > n + 500[/spoiler]

By flipping the second condition and adding it to the third condition, we get
<span class="math">28 a + 196 \leq 500[/spoiler]
<span class="math">28 a \leq 304[/spoiler]
<span class="math">a \leq 10[/spoiler]

By flipping the last condition and adding it to the first condition, we get
<span class="math">275 < 32 a[/spoiler]
<span class="math">8.6 < a [/spoiler]

So a = 9, 10

If a is 10, then n is between 82 and 100, because <span class="math">9^2 = 81[/spoiler], and because <span class="math">24^2 = 576[/spoiler], <span class="math">25^2 = 625[/spoiler], there's no danger of running into 25. This gives us 19 values of n.

If a is 9, then n is 65 to 75, because <span class="math">24^2 = 576 > n + 500[/spoiler], <span class="math">8^2 = 64 < n[/spoiler]. This gives us 11 values for n.

All in all, there are 30 values. To get the original values of n, add 250, to get 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350.

Remember, if there are 15 squares, there are 14 more than whichever is your reference. A lot of these analyses seem to miss that.

Sorry I couldn't post yesterday.[/spoiler]

If <span class="math">|n - m^2| \leq 250[/spoiler], then the range is 501 numbers; the squares lie between some number n and 500 more than that number. Because these numbers correspond one-to-one with the original n, I'll now only refer to the new number as n, and use the condition <span class="math">n \leq m^2 \leq n + 500[/spoiler]
So we have a number of conditions on the least a such that m = a solves this.

<span class="math">a^2 - 2 a + 1 < n[/spoiler]
<span class="math">a^2 \geq n[/spoiler]
<span class="math">a^2 + 28 a + 196 \leq n + 500[/spoiler]
<span class="math">a^2 + 30 a + 225 > n + 500[/spoiler]

By flipping the second condition and adding it to the third condition, we get
<span class="math">28 a + 196 \leq 500[/spoiler]
<span class="math">28 a \leq 304[/spoiler]
<span class="math">a \leq 10[/spoiler]

By flipping the last condition and adding it to the first condition, we get
<span class="math">275 < 32 a[/spoiler]
<span class="math">8.6 < a [/spoiler]

So a = 9, 10

If a is 10, then n is between 82 and 100, because <span class="math">9^2 = 81[/spoiler], and because <span class="math">24^2 = 576[/spoiler], <span class="math">25^2 = 625<span class="math">, there's no danger of running into 25. This gives us 19 values of n.

If a is 9, then n is 65 to 75, because 24^2 = 576 > n + 500, 8^2 = 64 < n. This gives us 11 values for n.

All in all, there are 30 values. To get the original values of n, add 250, to get 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350.

Remember, if there are 15 squares, there are 14 more than whichever is your reference. A lot of these analyses seem to miss that.

Sorry I couldn't post yesterday.[/spoiler][/spoiler]

If <span class="math">|n - m^2| \leq 250[/spoiler], then the range is 501 numbers; the squares lie between some number n and 500 more than that number. Because these numbers correspond one-to-one with the original n, I'll now only refer to the new number as n, and use the condition <span class="math">n \leq m^2 \leq n + 500[\math]
So we have a number of conditions on the least a such that m = a solves this.

<span class="math">a^2 - 2 a + 1 < n[/spoiler]
<span class="math">a^2 \geq n[/spoiler]
<span class="math">a^2 + 28 a + 196 \leq n + 500[/spoiler]
<span class="math">a^2 + 30 a + 225 > n + 500[/spoiler]

By flipping the second condition and adding it to the third condition, we get
<span class="math">28 a + 196 \leq 500[/spoiler]
<span class="math">28 a \leq 304[/spoiler]
<span class="math">a \leq 10[/spoiler]

By flipping the last condition and adding it to the first condition, we get
<span class="math">275 < 32 a[/spoiler]
<span class="math">8.6 < a [/spoiler]

So a = 9, 10

If a is 10, then n is between 82 and 100, because <span class="math">9^2 = 81[/spoiler], and because <span class="math">24^2 = 576[/spoiler], <span class="math">25^2 = 625, there's no danger of running into 25. This gives us 19 values of n.

If a is 9, then n is 65 to 75, because 24^2 = 576 > n + 500, 8^2 = 64 < n. This gives us 11 values for n.

All in all, there are 30 values. To get the original values of n, add 250, to get 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350.

Remember, if there are 15 squares, there are 14 more than whichever is your reference. A lot of these analyses seem to miss that.

Sorry I couldn't post yesterday.[/spoiler][/spoiler]

>>4370042
This means there are 16 integers that satisfy, not 15.

I like the double-counting. That's awesome.

>>4364688
Well, fuck. I didn't see that.

Have the counter flip the switches both down. Have each person flip switch 1 on the first time they see all the switches both down, and flip switch 2 up the second time they see both switches down. The counter has to flip the switches both down for someone to flip the second switch, so once the counter has 22 second-switch counts, everyone's been in the room.