Admiral Euphoria's Random Banter Thread of Delusional Grandeur #124

[wadledo]

Huh, that is pretty cool. I had never noticed before.

And the 3 and 6 tables always equal a multiple of 3, in case you didn't realize.

[Alteran]

And the 3 and 6 tables always equal a multiple of 3, in case you didn't realize.

But...how couldn't they? I mean, the 3 table is just multiples of 3. Multiples of 6 will also have to be divisible by 3 since 6 is. Maybe I don't quite understand what you mean?

[Thufir]

He means the sum of the digits of a multiple of 3 always equals a multiple of 3.

An interesting one: If you alternately add and subtract the digits of a multiple of 11, the answer is always 0.

[Dogmantra]

An interesting one: If you alternately add and subtract the digits of a multiple of 11, the answer is always 0.

That's pretty cool. I like the method of multiplying 2 digit numbers by 11: first digit, both digits added together, second digit, though if you get more than 10 when adding digits together, you add the first digit of the sum to the first digit of the original number.

EXAMPLE: 65 x 11 = 6 (6+5) 5 = 6115 715.

EDIT: Hang on... Thufir's trick doesn't work on this one!

[Alteran]

65 * 11 = 715

7 - 1 + 5 = 11

Hmm, yeah. Doesn't work.

[unstattedCommoner]

EXAMPLE: 65 x 11 = 6 (6+5) 5 = 6115 715.

EDIT: Hang on... Thufir's trick doesn't work on this one!

59 x 11 = 649 (oops)
60 x 11 = 660 (works)
61 x 11 = 671 (works)
62 x 11 = 682 (works)
63 x 11 = 693 (works)
64 x 11 = 704 (oops)

66 x 11 = 726 (oops)
67 x 11 = 737 (oops)
68 x 11 = 748 (oops)
69 x 11 = 759 (oops)
70 x 11 = 770 (works)

v: It does work if you add mod 11: 6+9 = 4; 7+4 = 0; 7+6 = 2; 7+7 = 3, etc...

[Pyrian]

What's amusing to me is that Dogmantra's example clearly demonstrates why it doesn't work in that case: it's because the 6 and the 5 exceed 9.

EDIT: Look at the line of examples: in each case where the sum of the two digits exceeds 9, you get an "oops".

This is because the effect depends on the fact that the two digits in 11 cancel each other out, and continue to do so in multiples, and continue to do so in 10's thereof as well (as a zero has no effect). But once the sums of two different 10's of 11 cause the next digit up to increment, the whole thing falls apart.

[Thufir]

Sorry, I remembered it wrong. They always add up to a multiple of 11.

[Pyrian]

That makes sense; by incrementing a digit you either gain one and lose ten or gain ten and lose one in the final number, making for a factor of eleven.

[HellfireLover]

I see lots of numbers, and now I feel stupid.

[Thufir]

And I'm not even trying!

[Player_Zero]

So... If you feel stupid about that then how about this?

Spoiler

Show

Let A be a number which is a multiple of three and the digits of which add up to a multiple of three.

A=3B where B is a natural number and a factor of A
3C=a₁+a₂+...+a_n where C is a unique natural number for each A we choose, a_t is the 't'th digit along the number A and where n is the total number of digits.

Then is we add three to A then clearly A is still a a multiple of three and then, by considering what happens to the final digit in the multiples of three we can conclude that eiter a_n is increased by three or decreased by seven and a_n-2 is increased by one. In either case there is still a natural number C that satisfies the equation, either it is one more or two less than before.

We can extend this, since we can just iterate adding threes; so adding any multiple of three to the number still satisfies it.

It works for three, so works for all multiples of three. Though technically you would have to explain why it works for negative multiples too.

It's just a proof of that 'any multiple of three has digits which add to a multiple of three' dealie.

I mean, it's a fairly simple inductive proof and I have long-winded it a bit, but it is pretty obvious...

[Dogmantra]

Sometimes, I like to have little conversations with people in PM titles, in my signature or in the "reason for editing" bit. I'm planning on bringing this into real life with a pad of post-it notes and a lot of free time. I hope people will be co-operative. This all ties in with my plan to make the world a better place, which I have titled "Project Awesome".

[unstattedCommoner]

So... If you feel stupid about that then how about this?

Spoiler

Show

Let A be a number which is a multiple of three and the digits of which add up to a multiple of three.

A=3B where B is a natural number and a factor of A
3C=a₁+a₂+...+a_n where C is a unique natural number for each A we choose, a_t is the 't'th digit along the number A and where n is the total number of digits.

Then is we add three to A then clearly A is still a a multiple of three and then, by considering what happens to the final digit in the multiples of three we can conclude that eiter a_n is increased by three or decreased by seven and a_n-2 is increased by one. In either case there is still a natural number C that satisfies the equation, either it is one more or two less than before.

We can extend this, since we can just iterate adding threes; so adding any multiple of three to the number still satisfies it.

It works for three, so works for all multiples of three. Though technically you would have to explain why it works for negative multiples too.

It's just a proof of that 'any multiple of three has digits which add to a multiple of three' dealie.

I mean, it's a fairly simple inductive proof and I have long-winded it a bit, but it is pretty obvious...

Unnecessarily convoluted.

Let A = a₀ + 10a₁ + 10²a₂ + ... + 10ⁿa_n for integers a_r, 0 <= a_r <= 9.

Now, 10 = 1 (mod 3). Therefore 10^r = 1 (mod 3), and

A = a₀ + a₁ + ... + a_n (mod 3).

So if A is a multiple of 3, then the sum of the digits of A is also a multiple of 3.

Similar proofs establish the tests for divisibility by 9 (10 = 1 (mod 9)) and
11 (10 = -1 (mod 11)).

One can obtain divisibility tests for all primes in this manner: it's just that they don't reduce to memorable rules like 2 and 5 (A is divisible if a₀ is), 3, and 11. 7, for example, yields nothing more useful than:

If a₀ + 3a₁ + 2a₂ + 6a₃ + 4a₄ + 5a₅ + a₆ + etc, continuing the
cycle, is divisible by 7, then A is.

Thus 7¹⁰ = 282475249:
1*9 + 3*4 + 2*2 + 6*5 + 4*7 + 5*4 + 1*2 + 3*8 + 2*2
= 9 + 12 + 4 + 30 + 28 + 20 + 2 + 24 + 4
= 133 = 19*7.

[Player_Zero]

Yes, yes, you're compelled to try to one-up.

[Cyrano]

Unnecessarily convoluted.

Let A = a₀ + 10a₁ + 10²a₂ + ... + 10ⁿa_n for integers a_r, 0 <= a_r <= 9.

Now, 10 = 1 (mod 3). Therefore 10^r = 1 (mod 3), and

A = a₀ + a₁ + ... + a_n (mod 3).

So if A is a multiple of 3, then the sum of the digits of A is also a multiple of 3.

Similar proofs establish the tests for divisibility by 9 (10 = 1 (mod 9)) and
11 (10 = -1 (mod 11)).

One can obtain divisibility tests for all primes in this manner: it's just that they don't reduce to memorable rules like 2 and 5 (A is divisible if a₀ is), 3, and 11. 7, for example, yields nothing more useful than:

If a₀ + 3a₁ + 2a₂ + 6a₃ + 4a₄ + 5a₅ + a₆ + etc, continuing the
cycle, is divisible by 7, then A is.

Thus 7¹⁰ = 282475249:
1*9 + 3*4 + 2*2 + 6*5 + 4*7 + 5*4 + 1*2 + 3*8 + 2*2
= 9 + 12 + 4 + 30 + 28 + 20 + 2 + 24 + 4
= 133 = 19*7.

Why the get out is this more confusing?

[Player_Zero]

Why the get out is this more confusing?

Because he used modular arithmetic, more symbols and didn't consider his audience?

[Cyrano]

Because he used modular arithmetic, more symbols and didn't consider his audience?

No, I think it's the font.

RIIIIIIIIIIIIIISK

[Player_Zero]

No, I think it's the font.

RIIIIIIIIIIIIIISK

No Risk. Not until you do your chores.

[Alteran]

Ooh, I might be up for Risk.

[Cyrano]

No Risk. Not until you do your chores.

I cleaned my room, took out the trash, made that mix tape for your big unexplained first date, did the dishwasher, put the new load in, sharpened your bread knife, checked the laundry for kitchen-related bloodstains, cleaned aforementioned bloodstains AND laid down the plastic wrap in the basement for your waterfight.

I think I've earned a little world domination.

EDIT: ^Whabam! How many do we need? This is good, keep it coming people, huptwoetc.

[Coidzor]

Modular arithmetic? What's that grouped in with, generally?

In other, more Coidzarian news (no word yet on whether Coidzarian or Coidzorian or Cuidan is really the proper way to write it out), I am heading out into the wilderness of _._. to release an Opossum female young adult that my brother found abandoned as a kit.

Wish me luck. Or for a drop bear to eat me. Though I think any drop bears in the U.S. would be very confused and starve to death or get eaten by cats/dogs/yuppies depending upon what area they settled in.

Hopefully I'll be able to unload some of this furniture on my bro or something or at least convince the 'rents to give it to goodwill to boot. But first.... plastic tubs...

Interestingly, I've stumbled onto something. Where previously a chain fit around my body and I could slip into or out of it with a little effort, though it would stay in place by virtue of my girth and friction around my midsection, now the chain merely falls straight to the floor.

So I think I've lost a bit to my waist haha.

[Alteran]

At least 3 for a decent game, but I prefer 4 or more. I may be able to dig up a 4th player, but probably not. He has work tomorrow, so no staying up late.

[Thufir]

Because he used modular arithmetic, more symbols and didn't consider his audience?

Alternatively because having read your proof, reading another seemed pointless. And because despite the fact he said yours was unnecessarily convoluted, his appears to be longer (Though that could be partly because he used normal sized text).

Or even, having actually read his proof now, because the actual proof is maybe a third of his post, the rest being taken up by related but off-putting looking extra stuff.

[Cyrano]

Alternatively because having read your proof, reading another seemed pointless. And because despite the fact he said yours was unnecessarily convoluted, his appears to be longer (Though that could be partly because he used normal sized text).

Or even, having actually read his proof now, because the actual proof is maybe a third of his post, the rest being taken up by related but off-putting looking extra stuff.

Sooo...that's a "yes" for Risk, then?

[Thufir]

Don't be ridiculous. I came here to talk about maths, thus continuing the RB time loop which amplifies my superpowers.
'sides, last time I played Risk it took 10 hours, so I'm hesitant to start a game at 1:30am.

[littlequietguy]

'sides, last time I played Risk it took 10 hours, so I'm hesitant to start a game at 1:30am.

I once played monopoly for 2 years. I lost.

[Alteran]

Don't be ridiculous. I came here to talk about maths, thus continuing the RB time loop which amplifies my superpowers.
'sides, last time I played Risk it took 10 hours, so I'm hesitant to start a game at 1:30am.

It just means you'll stay up really late to finish the game, and exhaustion will make you lose. I don't see the problem with this.

[Thufir]

It just means you'll stay up really late to finish the game, and exhaustion will make you lose. I don't see the problem with this.

I don't think Risk requires enough brainpower that I would be sufficiently impaired by exhaustion to make me lose if I was otherwise winning. But I'm disinclined to mess up my sleep pattern playing Risk over the internet.

[Coidzor]

Ahh, endurance games. Hmm...

What do you do when your hat starts to smell like your dog and is developing a fine coating of lint and cat hair?

I'm trying to look for a lint roller, but I really don't wanna febreeze my hat anytime while it would be in ocntact with mein head.