Eighteen

[rbandrews]

The number 1818 is a number, 18, repeated twice.

If you add together the digits, 1+8+1+8, you also get 18.

There is no other number with this property.

Comments

[desfido.livejournal.com]

What exactly are you considering to be "this property"? Since, for example, I'd say that what you showed is a consequence of the following property which I conjecture: if you have 9y, and repeat the digits of 9y y times, the sum of the digits is 9y, for either any natural number y (using the "0 is not a natural number" version). If I get bored enough waiting to see somebody at the health center later, or in one of my classes, I may go ahead and see if I can prove that, in fact, or figure out what is special about the natural numbers for which it is true (since it's clearly true for a bunch of them).

Anyway, point is, what specifically is the property which you are asserting 1818 uniquely holds? The main contender I can think of is equivalent to the less interesting "1000x+100y+10x+y=1818 and 2x+2y=10x+y, if and only if x=1 and y=8".

[desfido.livejournal.com]

You can tell I'm sick from the dumb mistakes in the bit after the colon, but you get the idea, I figure.

[gigaclon.livejournal.com]

yes this is true, but only in base-10 in base-16 1818 adds up to 6 and the number 1818 is impossible in base-2 and base-8

[rbandrews]

It's the only number like that where the sub-number (18) is repeated twice.

There are a bunch of numbers if you're allowed to repeat is more than twice, which I'm going to call demure numbers for lack of a better adjective and talk about more this evening.

[rbandrews]

Also, it is possibly true for every multiple of nine, but it's not only true for multiples of nine. 24 is demure, as is 84. And 10.

[wheeerdball.livejournal.com]

The proof for why this is the only answer can easily be re-constructed for alternative basis. The part about how you can only have two digits in the sub-number applies to all basis n > 2. From there, the number is then uniquely determined to be 1 (n-2) 1 (n-2).

If the basis is n = 2, then the proof needs to be modified to state that there can be no more than three digits. But the number 1 0 1 0 still works, and by trial and error it can still be solved to show that there is no other answer.

[rbandrews]

In hexadecimal, 1E1E works.

[desfido.livejournal.com]

Okay, interesting, though irrelevant to my conjecture, since I didn't claim that only multiples of 9 would be "demure", merely that the multiples of 9 all were (didn't actually get around to proving this yet, btw, so still just a conjecture, unless you've proven it or seen the proof).

I fail to see why it's all fluffy and special that 18 is the only one whose demure number is 2. Maybe that's because I'm just not that into number theory, relative to other areas of math (probably mostly because I haven't done much of it). But, basically, why do I care that 18 is the only number which is demure for 2 (perhaps this will be addressed by your later post)?

[rbandrews]

Beats me. I found it interesting, so I wrote a post about it. It's not like everything needs a purpose or anything.

[desfido.livejournal.com]

Fair enough. And I suppose that's not exactly what I'm getting at, anyway. Basically, is there a specific reason you find the specific demureness of 18 in this way more interesting than demureness in general, or some other reason why you started out by writing a post about 18's demureness instead of demureness in general, or did you just think of / stumble upon the 18 one, post about it, then think of / stumble upon the others, or what?

[rbandrews]

Yesterday I drove by 1818 Main Street, which I reflexively added together and got another 18, which made me happy. Then I noticed that it was the only four-digit number like that, and started thinking about the general property of numbers like that, and started calling the property of numbers that you can use to construct numbers like that "demureness", and started annoying Aaron in the global marketing strategy meeting this morning with it.

But I found 1818 interesting particularly because it was the first one.

[rbandrews]

Yeah, you're right, I want to total 14, not have the second digit be 14. My mistake.

[wheeerdball.livejournal.com]

Good luck finding a pattern for the multiples of 9 where it works. Cases where it works/doesn't:

1-10: yes
11: no (9 + 9 = 18 = even, 99 = odd)
12-20: yes
21: no
22-30: yes
31: no

...

101: no
102: yes
103: no
104: yes
105: no
106: yes
107: no
108: yes
109: no
110: yes
111: yes
112 - 120: yes
121: no

...

201: no
202: yes
203: no
204: yes
205: no
206: yes
207: no
208: yes
209: no
210: yes
211: no
212: yes
213: no

[desfido.livejournal.com]

Huh. Fair enough. Due to some annoyances with various geology assignments, I haven't had time to get bored enough to investigate the conjecture yet. Offhand, though, I already think I see a pattern there, it's just an annoying one. When I actually have time (hopefully next week, but more probably not till after finals end on May 15 or so), I'll see if I'm right or just fooling myself.

Yes, I know the implication of your comment is that I'm probably just fooling myself, but I have to try it for myself. If I didn't trust my mom that the iron was hot when I was 3...

[rbandrews]

I noticed the 99 one before, but I apparently forgot to say anything.

Note that all multiples of 3 are demure. It's the test for whether a number is divisible by 3.