Post by: Nagisa on December 15, 2012, 10:21:23 pm
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.
Post by: TrueTears on December 15, 2012, 10:29:45 pm
Post by: FlorianK on December 15, 2012, 10:33:46 pm
A palindrome is a number or word that is the same when read forward and back- ward, for example, "176671" and "civic."I'd say no.
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.
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?
Post by: kamil9876 on December 15, 2012, 10:36:48 pm
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
Post by: kamil9876 on December 15, 2012, 10:38:57 pm
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
Post by: Nagisa on December 15, 2012, 10:43:49 pm
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.
Post by: brightsky on December 15, 2012, 10:50:05 pm
sorry got bored
perhaps one could get somewhere by considering the nature of the 'middle digit'.
Post by: FlorianK on December 15, 2012, 10:52:08 pm
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=111111987654321I never assumed that n=987654321
I used n=987654321 to prove that it is not possible for XXX(any number before)987654321
Post by: TrueTears on December 15, 2012, 10:52:44 pm
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
Post by: brightsky on December 15, 2012, 10:54:52 pm
Post by: kamil9876 on December 15, 2012, 10:57:24 pm
edit: exactly what brighstky said.
Post by: kamil9876 on December 15, 2012, 11:00:57 pm
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?
Post by: Planck's constant on December 15, 2012, 11:19:06 pm
I am not sure if there is a solution available anywhere
Post by: kamil9876 on December 15, 2012, 11:24:10 pm
Quote
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?
Post by: Nagisa on December 15, 2012, 11:27:28 pm
Of course it is, right in this thread. Unless you are talking about the base 2 case?
haha
Post by: Planck's constant on December 15, 2012, 11:28:48 pm
Of course it is, right in this thread. Unless you are talking about the base 2 case?
You are too fast for me :)
Post by: Nagisa on December 15, 2012, 11:37:34 pm
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.
i dont fully understand what you mean by this, are you saying that it cant be a palindrome because of the way the middle would turn out?
Post by: kamil9876 on December 16, 2012, 12:35:05 am
case1: The 10^k string is in the middle
case2: It is somewhere to the left
case3: It is somewhere to the right
The idea is to get a contradiction for each case. (yeah so you have to sorta read both my first and second post, the second one was a continuation of the first so yeah it doesn't make much sense if you read it on its own)