Maths Problem

[Nagisa]

A palindrome is a number or word that is the same when read forward and back- ward, for example, "176671" and "civic."
Can the number obtained by writing the numbers from 1 to n in order (for some n > 1) be a palindrome?

make sure to post even basic ideas lol, don't be afraid.

[TrueTears]

well it seems the last few digits of the number needs to be at least 321 for it to be a palindrome

[FlorianK]

A palindrome is a number or word that is the same when read forward and back- ward, for example, "176671" and "civic."
Can the number obtained by writing the numbers from 1 to n in order (for some n > 1) be a palindrome?

make sure to post even basic ideas lol, don't be afraid.

I'd say no.
The start is 123456789
So the end of the palindrome has to be
987654321
This only happens when it is actually the number 987654321 XXX987654321
Lets take the the one without a number infront.
so n=987654321
the the number before is 987654320
however the start is 12345678910 so there is no 0 after the 9 in the start of our series, since for all XXX987654321 we would need a 0 after the first 9 numbers, but we have a 1 your assumption can't be true.

Does this make sense?

[kamil9876]

If you start from 0, and work in base 2 then you can

I reckon you can't, a sketch which hopefully someone can turn into a proof: Suppose it worked for 1,2, ... n  Then take the last number of the form in this sequence. It has k 0's. Hence our supposed palindrome has a long string of zeros on the right side, but on the left side it doesn't have such a string of zeros. Of course it may happen that this long string of zeros is in the middle, in that case there's another contradiction one can obtain.

[kamil9876]

I'd say no.
The start is 123456789
So the end of the palindrome has to be
987654321
This only happens when it is actually the number 987654321 XXX987654321
Lets take the the one without a number infront.
so n=987654321
the the number before is 987654320
however the start is 12345678910 so there is no 0 after the 9 in the start of our series, since for all XXX987654321 we would need a 0 after the first 9 numbers, but we have a 1 your assumption can't be true.

Does this make sense?

We don't need n=987654321, all we can conclude is that the last few digits of n are 987654321 i.e what if n=111111987654321

[Nagisa]

We don't need n=987654321, all we can conclude is that the last few digits of n are 987654321 i.e what if n=111111987654321

yeah i believe its possible, but yeah, i always tend to play around with algebra. maybe that will tender some result.

[brightsky]

123456789101112131415161718192021222324252627282930...039282726252423222120291817161514131211101987654321
sorry got bored
perhaps one could get somewhere by considering the nature of the 'middle digit'.

[FlorianK]

We don't need n=987654321, all we can conclude is that the last few digits of n are 987654321 i.e what if n=111111987654321

I never assumed that n=987654321
I used n=987654321 to prove that it is not possible for XXX(any number before)987654321

[TrueTears]

let us assume some bound for n. if this palindrome exists, it needs to at least end with 321, so assume that where k => 2.

now building on kamil's idea the longest consecutive sequence of 0's in our palindrome will be k digits long and these sequences will occur at multiples of 10^k

now... maybe you can try counting how many groups of consecutive 0's of length k there are and see if these groups of consecutive 0's can be "mirrored" onto each other when you reverse the digits of the palindrome, i have a feeling it's impossible

[brightsky]

i'm thinking, somewhere on the RHS of the ellipsis (in which there must be a 'middle digit' if there exists such a palindrome), there will be a string of 9's (i.e. 999999999999999), and the same string of 9's can't be present on the LHS.

[kamil9876]

Yeah in that case, there is only one substring "10000..00" with k zeroes. And that is precisely the one comming from that number . This is because a number cannot start with 0, yet it can't have more than k+1 digits. So the only way our number can be a palindrome is if the "100..00" substring is in the middle, but this gives a contradiction since walking a little bit to the left we get a lot of 9's, which are not found if you go a little bit to the right.

edit: exactly what brighstky said.

[kamil9876]

I guess this approach works for bases >2  (since we are essentially arguing with three different digits, 0,1,9 )

How about for base 2? Of course if you start from 0 then it works in base 2 with n=2 i.e  0110   But what if in base 2 you start from 1?

[Planck's constant]

This is problem 17 from the 22nd All Russian Mathematical Olympiad 1996.
I am not sure if there is a solution available anywhere

[kamil9876]

I am not sure if there is a solution available anywhere

Of course it is, right in this thread. Unless you are talking about the base 2 case?

[Nagisa]

Of course it is, right in this thread. Unless you are talking about the base 2 case?

haha