Current Events > A divisibility by 11 problem you've probably never seen.

(+) Yeah! (+)

Define s(n) as follows: s(1)=1 s(2)=12 s(3)=123 ... s(10)=1234567890. For n>10 the digits merely repeat. So for example s(13)=1234567890123.

Now the problem. What is the smallest natural number n such that s(n) is divisible by 11? What is the second smallest? And no calculators permitted as that defeats the purpose.

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no u

https://gamefaqs.gamespot.com/a/forum/0/050857bb.jpg

I passed college algebra and after that, I'm finished with any math I can't do on a calculator.

21 and 38

x

SkylightNight posted...

21 and 38

My spider senses agree

FLOUR posted...

Define s(n) as follows: s(1)=1 s(2)=12 s(3)=123 ... s(10)=1234567890. For n>10 the digits merely repeat. So for example s(13)=1234567890123.

Now the problem. What is the smallest natural number n such that s(n) is divisible by 11? What is the second smallest? And no calculators permitted as that defeats the purpose.

Are you doing a number theory PhD or something?

What does ChatGPT say!

SkylightNight posted...

21 and 38

agreed, although I'm not sure if there's a better way to get it after knowing the rule than just sorta sequentially checking them

Turbam posted...

What does ChatGPT say!

after figuring it out myself, I pasted the topic question into chatgpt and it came up with a(n actually accurate) method, explained the method in detail, started using the method to check each value of n one by one, and then stopped at 20 basically saying "now you know how to do it, it might take some time but you can continue this and find the answer" when it was literally 1 away from the first correct answer. a valiant attempt but also lol

What in the actual fuck is that?

Well, I said 'brown'.

Fifty...six...ish?

No

This pattern will occur repeatedly forever base on the TCs statement. The pattern will follow this many required increases to N from the previous hit to find a satisfactory number divisible by 11.

+21, +17, +5, +13, +9, +9, +13, +5, +17, +1 <repeat from start>

S(21), S(38), S(43), S(56), S(65), S(74), S(87), S(92), S(109), S(110) are all divisible by 11.

Now the sequence repeats:
(Note S(131) is +21 from 110.

S(131), S(148), S(153), S(166), S(175), S(184), S(197), S(202), S(219), S(220) are all divisible by 11.

Next repetition starts with S(241).

eaglei3 posted...

This pattern will occur repeatedly forever base on the TCs statement. The pattern will follow this many required increases to N from the previous hit to find a satisfactory number divisible by 11.

+21, +17, +5, +13, +9, +9, +13, +5, +17, +1 <repeat from start>

S(21), S(38), S(43), S(56), S(65), S(74), S(87), S(92), S(109), S(110) are all divisible by 11.

Now the sequence repeats:
(Note S(131) is +21 from 110.

S(131), S(148), S(153), S(166), S(175), S(184), S(197), S(202), S(219), S(220) are all divisible by 11.

Next repetition starts with S(241).

How the fuck do you figure this out without a calculator

PurestProdigy posted...

How the fuck do you figure this out without a calculator

The difference between s(1) and s(2) is 11, so the remainder from s(2) is the same as the remainder from s(1) (1 / 11). Between s(2) and s(3) is 111. That is one above a number divisible by 11 (110), so the remainder increments by 1 and is now (2 / 11).
1111 is divisible by 11 (1100 is divisible by 11, add 11 to that) so the remainder is the same as before.
11111 is 1 more than a number divisible by 11 (11110), so the remainder is now (3 / 11)

Work into the pattern the zeroes from s(10) and s(20) and you'll be able to work your way up to where the remainder is zero, giving you a whole number.

[LFAQs-redacted-quote]

There's a sort of pattern to the divisibility, I noticed it when I was trying to figure it out and thought it was so cool that I googled it and it turns out it's just normally well known (well, of course it is lol there's no way I'd discover something new like that)

I was multiplying random things by 11 to see if there was some kind of trick
https://gamefaqs.gamespot.com/a/forum/5/56efafe1.png
and you see you get the same number just sort of offset, so you can sort of match them up in pairs, each number with its neighbour. and they would cancel out perfectly in the end result except that sometimes you have to carry a digit when you get something over 10
https://gamefaqs.gamespot.com/a/forum/a/a2aeaebe.png
So basically when there is no carrying, any digit is fully balanced / cancelled out by its neighbours. which means you should be able to add up all the even place digits and subtract all the odd place digits and have them balance out to 0. Except again, there's carrying. And every time you carry you decrease your number by 10 (since it's like a 10 being written as a 0) and increase your neighbour's by 1, leading to a total difference of 11.

which means you can do the adding thing again and check for either 0, or some multiple of 11
https://gamefaqs.gamespot.com/a/forum/b/bf06e29a.png

I have no idea if this is the best way to do it but I dunno I noticed it and thought it was cool.

edit: then you just check the numbers as you go up. the digits tack on the end in a pretty structured way so it's not too hard to count ahead and just see where you'll be

PurestProdigy posted...

How the fuck do you figure this out without a calculator

I don't know how you do this with a calculator a normal way. The numbers get way to big.

I did it by using the addition and subtraction method for finding numbers divisible by 11.

Basically alternate adding and subtracting numbers and if the result is divisible by 11 or equals 0, you know that number is divisible by 11.

For instance.
S(21) = 123456789012345678901
S(21) is 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 0 +1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 0 +1 = 11

So S(21) is divisible by 11.

Now there are other patterns to recognize. The TCs problem repeats 1234567890 over and over. So for every S(+10), my overall total goes up by 5. Then using the numbers above, my string from the start always changes by:

+1, -1, +2, -2, +3, -3, +4, -4, +5, +0

So for figuring out S(38), I can shortcut if further. I know that S(30) is 15. Doing the simplified string above, I know that 15 - 4 is 11 again. -4 is the 8th number of the string above so I know that S(38) is divisible by 11.

As a random one picked out, I will come off of S(300). Since I know that every S(10) sequence drives my count number up by 5, that puts me at 150. I know from my multiplication table that 11 * 13 = 143 and 154. 150 is between that. Using the pattern, I know +4 from 150, which is S(300+7) more from will get me to 154. So that means S(307) is divisible by 11. If I want to go back to find what was divisible 143. I know to go back to S(280) which puts me at 140. + 3 on that is the 5th number, so I know that S(285) is divisible by 11 as well.

Please note that I can only do this second part because the TCs numbers follow a simple repeating pattern.

Yeah I messed up my math. I should have written it out not tried to do it in my head.

Still waiting for TCs response on why he keeps asking these weird math olympiad / number theory type questions on a video game website

Every day he posts one of these topics then disappears

What a weird schtick

[LFAQs-redacted-quote]

Close. Its 42

kirbymuncher posted...

agreed, although I'm not sure if there's a better way to get it after knowing the rule than just sorta sequentially checking them

after figuring it out myself, I pasted the topic question into chatgpt and it came up with a(n actually accurate) method, explained the method in detail, started using the method to check each value of n one by one, and then stopped at 20 basically saying "now you know how to do it, it might take some time but you can continue this and find the answer" when it was literally 1 away from the first correct answer. a valiant attempt but also lol

Actually, that sounds great. If chatgpt comes with a 'homework detected' filter that makes it show you how to solve the problem and avoids just handing out the answer, I absolutely have no complaints.

I don't think it's homework guys, it gives me vibes more of like a math contest problem for 10 year olds

Daremo posted...

Actually, that sounds great. If chatgpt comes with a 'homework detected' filter that makes it show you how to solve the problem and avoids just handing out the answer, I absolutely have no complaints.

I don't think it's anything that complex, I think 20 is just a nice round number to stop at especially when it has response length limits, and it's just a coincidence that the answer is 21. bad luck on its part

that said, I dunno if you've asked it anything recently (or at all) but compared to last year or even just months ago, chatgpt is an extreme overexplainer right now. Unless you very firmly tell it multiple times to not treat you like an idiot it will write annoyingly excessive amounts of explanation/preface to even simple questions

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