IQ (moved subtopic)

the property is true for alll numbers n divisible by 3
for numbers 3n+1 you can get the sum of the digits just by adding 1 to the sum of the digits for 3n unless 3n ends with 9.

only problem is if it ends with 9 but the property is also true for 3n+3 which now ends with 2
so we can get the sum of the digits of 3n+1 by subtracting 2 from the sum of the digits of 3n+3

in both case it can be written as 3p+1 or 3p-2 from some p so it can't be divisble by 3

similarly for 3n+2

Third one is a trick question. You've obviously never heard of...

Not going through all that, it's probably correct, but as I say, a bit too cumbersome for my liking. Sklansky's original proof was simple enough to explain to an 8 year old but incomplete. The patched proof has lost that element of simplicity as we now have to deal with remainders, so it has no benefit over the general proof.

But then it has no answer lol.

It's almost too simple to write down. Maybe easier to see with an example

15 is divisble by 3. Sum of digits = 6 which is divisible by 3
16 = 15+1. sum of digits = 6+1. not divisble by 3

99 is divisble by 3. sum of digits =18 which is divisble by 3
99+3 = 102. sum of digits = 3 which is divisble by 3
100 = 102 -2 , sum of digits = 3 - 2. Not divisble by 3

all 3n+1 can be done that way. Similarly for 3n+2
(assuming the proof for n divisible by 3)

Gotcha!

Not sure your "trick" is as clean/intuitive as you think, try it on:

2999999997
2999999998

Then you still need a step to demonstrate that we have accounted for all the natural numbers in this argument, otherwise the same hole remains.

But yeah, it's basically what I suggested in my follow up post. Just not a fan, I don't see what benefit it has over the general proof, and it doesn't prove nearly as much.

And don't call me Shirley.

sum of digits of 2.......8 = sum of the digits of 2.......7 + 1
the last digit is incremented by 1 the rest stay the same. So only the last digit effects the sum and it's gone up by 1

yeah all the natural numbers are 3n, 3n+1 or 3n+2

I definitely prefer your 'static' proof. Was just filling in the gap you pointed out

It works, but it's just not a good proof. If all you had to go by was this proof and nothing else, you'd be inclined to think this was some weird unexplained coincidence with just the number 3 and base 10, none of which is true.

It's a perfectly good proof of the gap you identified

It doesn't provide any intuition as to what is going on afaics

So chez, how many IQ points are we subtracting from Sklansky for this error in basic logical reasoning (while solving a problem he himself suggested as a test of IQ, no less)? What did he start at, 200? I say we dock him 80 points and let him work his way back up.

Or maybe you're significantly better at math than even the best HS student in New Jersey? Makes sense since everyone here is at least 2 SDs above the mean.

I'm not sure ignoring the uninteresting bit is a logic fail.

If pointing out the omitted potential counter arguments to DS's posts is a measure of high IQ then hello there!

slurp

Saying you proved "if x then y" whereas what you actually proved is "if y then x" is a spectacular and complete logic fail. The proof being incomplete is not the fail. Failing to recognise what you proved is the fail.

Surely a simpler way of looking at is that the position of a digit in the number is irrelevant to its modulus when divided by 3. 8, 80, 800, 8000 all have a modulus of 2. For 7 and 700, it’s 1. So by adding the digits you are also adding the mods of the digits which will either exactly divide or not.

Yeah wete going to disagree but whatever makes you feel good

I didn't want to be so rude but DS has more of an engineer mindset than that of a pure mathematician. There I said it.

What's he "engineering" if he can't solve basic arithmetic problems, lego houses?

The bolded is just a restatement of the problem, it's not a proof. That is the statement which needs to be proven. Why is the position of a digit irrelevant to the modulus of the number when divided by 3 but not when divided by 7? 8, 80 and 800 all have different remainders mod 7.

You're reaching a conclusion your logic can't support.

Why you need to feel so superior is odd. You're vert decent at maths, you're logic and analysis ain't so hot at times.

Hopefully it's a bit better than your spelling and grammar.

Depends what you mean by 'better'?

English teacher would be prouder of you.

i mean your worser

You're catrching up.

All numbers can be reached by continually adding 3 to the numbers one, two or three. If you do that to ANY multiple digit number, you are always gonna add three or subtract seven from the last digit while adding one to the next to last digit, or occasionally subtract seven and turn consecutive nines to zeroes before adding one.

In any case you are adding or subtracting a multiple of three from the original number when you add three to the digit sum.

But if you add or subtract a multiple of three from a number that is not a multiple of three it obviously won't result in a multiple of three.

But all non multiples of three can be reached by adding three at a time from a start of one or two. And all that started with three have digit sums that are multiples of three. So that takes care of all numbers. There can't be a weird number outside of these categories that somehow has its digits add to a three multiple when the number doesn't.

I sometimes overlook steps because they seem obvious.