all i see here is someone desperately wanting a high sklanksy rating

it isn't necessarily absurd no.

all it takes is to have tasks that you define as requiring intelligence and where you can measure outcomes in a strictly rankable way.

then you define AGI, for ex the definition I prefer is:

AGI would generate outcomes in all such tasks that are better than what the best human result at that ever was, and with intervals of time, AGI at t x+y will generate outcomes that are better than AGI at t x.

so you get generalized super human intelligence that self improves.

I prefer that definition because that should be want brings the singularity about

The video is aware of both the cartridge and itself. The cartridge is aware of neither.

people are far too attached to static solutions in dynmaic systems but this is definitely a static type problem imo.

another 'dynamic' approach is inductive, if it's true for a number n digits long then it's obviously true for a number of n+1 digits long

Judging by how free-flowing the points have been so far, that's got to be worth at least 160.

You might want to think about this some more. Hint: a number 1 digit long is divisible by 7 if its digits add up to 7. The number 16 is 2 digits long and its digits add up to 7.

I think you meant "if it's true for a number n digits long then it's obviously true for a number n - 1 digits long" which, while correct, does not help us with the inductive approach, since you have no useful base case.

Yeah I have no idea what you're talking about as usual, but it looks totally wrong.

Number of digits k? It's true for divisibility by 7 for k = 1. It's not true for divisibility by 7 for k > 1.

It is true for all k for divisibility by 3, but this has already been shown in general for any k, so it has nothing to do with induction.

I may have to rethink it

If you have a number that is both divisible by three and whose digit sums are divisible by three then if you add three it is usually instantly obvious that the new number will retain both descriptions. If xxxxxx5 works so will xxxxxx8.

It is slightly less obvious if the right hand digit is seven, eight, or nine If it is one of those three digits then adding 3 will (usually) reduce that last digit by 7 but add one to the next to last digit. The reduction of the digit sum will be six, so the full number will retain both traits. xxxxx58 becomes xxxxx61.

The exception would be if the original number not only ends in 7, 8, or nine, but the digits to its immediate left are one or more consecutive nines. In that case when you add three, the last digit reduces by seven, the nines all flip to zero and the digit to the left of the nines goes up by one. abc99998 turns into [ab (c+1)00001. Subtract seven, subtract a bunch of nines and add one and its still obviously a digit sum divisible by three. (Note: in the case of a number consisting of only nines you would declare the digit to their left as being zero. 9999 is 09999)

Since you can start this process at the very beginning, namely the number "three" and since adding three at a time will give you all numbers divisible by three, then we can say that a number will be divisible by three if and only if the sum of its digits are.

That works to explain the specific case exact divisibility by 3 in base 10, but it doesn't generalise vey easily. With the other explanation you can see the solution in full generality, for all divisors for which this works (3 and 9), all remainders on division by those numbers (not just remainder 0) and for all bases (not just base 10).

In a given base b, when you add all the digits of a number together what is left over to get to the original number is b-1 multiplied by some stuff. That is all going to be 0 mod b-1 or any of its divisors. Therefore the sum of the digits mod b-1 or any of its divisors will be the same as the original number mod that divisor.

But yes, your explanation would be the induction proof that chez was looking for.

yes the induction on n to n+1 digits works but it requires DS step so it's a tad pointless

unless we can say DS step counts as obvious in which case it's correct as I stated

Why would the video be aware but not the cartridge? Both have electrical signals running through them. Why would light coming out of a screen make a difference between them?

Wait a minute, what do you mean by video?

That makes no sense at all: while there is a clear correlation between scoring high on an IQ test and solving math problems and/or having a way with words there is no such correlation for IQ tests and drawing realistic paintings, at least it's not the consensus. I fail to see the point you're trying to make.
My hypothetical involves one widely accepted correlation and yours doesn't.
My point is IQ tests do measure a set of skills. If you at least tried to answer honestly you'd prove me right.

A person on the bottom left of IQ distribution won't necessarily know what "realistic" means in the context of a painting. IQ tests can tell you someone has major issues (when they test very far to the left of the median of their population)

Well he or she may not know what realistic is but draw a painting that reflects the concept. Anyway I'd go with someone who's on the bottom right to do it (in the hypothetical it was either bottom left or the opposite).

I'm surrounded by Philistines.

Just because they got the lowest IQ in Asia you gotta be racist now?

The ability to logically reason out what aspects of a painting make it look realistic, what kind of paint supplies and drawing styles are conducive to realistic painting, and guessing the possible reactions a person would have to various paintings you could choose to attempt to draw, would all be aided by having a high IQ, and impaired by having a low IQ. So, all other things being equal, of course you would favor the high IQ individual. There would have to be a significant negative correlation between high IQ and painting ability to do otherwise, since you're giving up so much reasoning ability that is useful to the task.

It would be more interesting if the choice was between a low IQ professional painter and a high IQ skilled amatuer or semi-professional painter.

Actually, I don't think this is even a proof. You have shown that all multiples of 3 have this property, since starting with 3 we can reach all multiples of 3 and demonstrate that they have the property. But you haven't shown that all numbers with this property are multiples of 3. There is nothing in this proof that precludes a random number that is not a multiple of 3 having a digit sum which is 0 mod 3.

Therefore, this does not prove that a number with a digit sum which is 0 mod 3 is a multiple of 3. It proves that if a number is a multiple of 3 it has a digit sum which is 0 mod 3. The statement "a number is divisible by 3 if the sum of its digits is" is actually not proved. The statement that is proved is "if a number is divisible by 3, then the sum of its digits is".

To complete this proof you also need to show that no numbers other than multiples of 3 have this property, but I'm not sure how you'd do this in the spirit of this proof without falling back on the more general proof anyway.

I suppose you could make a similar inductive argument for digit sums 1 mod 3 and 2 mod 3 then argue that since we've accounted for all possible remainders, all the natural numbers are accounted for, although that really seems to be getting into the territory where the general proof is much easier to follow. But if you were going to go down that route, once you have proved these statements:

if a number is 1 mod 3 its digit sum is 1 mod 3
if a number is 2 mod 3 its digit sum is 2 mod 3
if a number is 0 mod 3 its digit sum is 0 mod 3

then both the "if" and "only if" predicates of the statement in your proof would be satisfied.

For all n divisible by 3 : sum of digits of n = 3p for some p

for integers = 1 mod 3
if last digit isn't 9 then sum of digits = 3p+1 for some integer p
if last digit is 9 then sum of digits = 3p-2 for some integer p

similarly for integers 2 mod 3

slowish pony

Alright. I have an exercise for everyone. What letters come next in each of the sequences below?

P H L V ?

P H L H L L M T ?

P H L L ?

I M R T R D ?

This doesn't prove anything, it just states it. Also, "3p+1 for some integer p" and "3p-2 for some integer p" are the same thing, they both mean 1 mod 3. By convention, you say "3p+1" and choose p accordingly.

You could follow the same inductive argument as Sklansky's to show that all the remainders obey the rule, but I think it just becomes far too cumbersome to be of any benefit and you might as well use the general proof.