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My 1000th post.

Maybe I should give my keyboard a rest.

Anyway Happy Easter, and here's a little gem for you.

Take a number N, reverse its digits and add it to N.

Repeat if necessary. Eventually you will get a palindrome.

For example, starting with 39, we have 39 + 93 = 132.

Then 132 + 231 = 363 = palindrome.

After a moment's thought, this is not surprising.

But sometimes you do have to be patient.

Consider the starting number 89:

```					  89	  ------>   159487405
187	 |		  664272356
968	 |		 1317544822
1837	 |		 3602001953
9218	 |		 7193004016
17347	 |		13297007933
91718	 |		47267087164
173437	 |		93445163438
907808	 |	   176881317877
1716517	 |	   955594506548
8872688	 |	  1801200002107
17735476	 |	  8813200023188  =  palindrome!
85189247 --->[/code]```

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My 1000th post.

Maybe I should give my keyboard a rest.

Anyway Happy Easter, and here's a little gem for you.

Take a number N, reverse its digits and add it to N.

Repeat if necessary. Eventually you will get a palindrome.

For example, starting with 39, we have 39 + 93 = 132.

Then 132 + 231 = 363 = palindrome.

After a moment's thought, this is not surprising.

But sometimes you do have to be patient.

Consider the starting number 89:

```					  89	  ------>   159487405

187	 |		  664272356

968	 |		 1317544822

1837	 |		 3602001953

9218	 |		 7193004016

17347	 |		13297007933

91718	 |		47267087164

173437	 |		93445163438

907808	 |	   176881317877

1716517	 |	   955594506548

8872688	 |	  1801200002107

17735476	 |	  8813200023188  =  palindrome!

85189247 --->```

HAPPY EASTER and HAPPY 1000th POST DAY BONONOVA.

And what a mind-boggling post that is.

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HAPPY EASTER and HAPPY 1000th POST DAY BONONOVA.

And what a mind-boggling post that is.

Nice post, but I would be amazed if anyone can prove this!!

There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.

For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.

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Nice post, but I would be amazed if anyone can prove this!!

There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.

For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.

Hi neida.I just wanted to compliment bononova on his 1000th post.You bet I don't intend to prove it .Atleast not now.But your challenge is definitely an interesting one . Maybe I can prove the contrary. But putting all maybe's aside , please post your answer on The palindromic puzzle of yours . It is frustrating to know the question and not the answer for 2 whole days.So please post your answer soon.Sorry for the initial lack of courtesy.(forgot to include please).

Edited by grey cells

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Nice post, but I would be amazed if anyone can prove this!!

There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.

For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.

Wow!

I was looking for some proof by induction. I found out that some numbers can be expressed as sum of two numbers, n and n', such that n' is obtained by revesing the digits of n, and others which cannot be expressed as such. Both classes contain palindromes. 111 is a palindrome which cannot be expresses as sum of n and n'.

However, the above post has helped me in deciding to abandon this problem now.

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Nice post, but I would be amazed if anyone can prove this!!

There are actually some numbers that are thought never to resolve to a palindrome in this way - 196 is the smallest example. These numbers are called "Lychrel numbers". However, no one has ever proven that they exist, simply no one has managed to prove that they don't.

For example, 196 has been processed through supercomputers, but even when the resultant number is over 300 million digits long, it still isn't a palindrome.

Neida, I am not a Math major, the last time I read Math was almost 17/18 years ago. My knowledge rotates mostly around common sense.

I looked up what "Lychrel numbers" were and it showed up on Wikipedia. With respect to 196, they seem to stick with the defenition of how to rotate numbers and get palindromes, but with the rest they dont seem to. Can you explain why? (As i said, i am not a math major)

`A Lychrel number is a natural number which cannot form a palindrome through the iterative process of repeatedly reversing its base 10 digits and adding the resulting numbers.`

Example:

```
# 56 becomes palindromic after one iteration: 56+65 = 121.
# 57 becomes palindromic after two iterations: 57+75 = 132, 132+231 = 363.
# 59 is not a Lychrel number since it becomes a palindrome after 3 iterations: 59+95 = 154, 154+451 = 605, 605+506 = 1111[/codebox]```

Now their Example for 196:

[codebox]
196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, 1997

Should it not go like this instead?

` 196 887 1675 7436....`

I know it amounts to the same, but not having studied Abstract Math, I do not understand this. please explain or point to a good source to read up on this.

Thanks. (PM me if u need my email address so we can take this offline. Thanks)

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Why would you take this offline? The whole point is to discuss it here

my idea:

if the number is made up of digits that in total add up to less than 10 (ie, add up to a single digit), than you get it in the first round:

2+1 = 3, which is a single digit, so:

21 + 12 = 33

4+1+1+2+1 = 9, single digit, so:

41121 + 12114 = 53235

1117 adds up to 10, but it doesnt necessarily mean it WONT be a palindrome on the first round:

1117 + 7111 = 8228

1218 + 8121 = 9339

So that leads to a more general (and obvious) rule, that if the corresponding flip-pairs each add up to a single digit, there is a palindrome in the first round, because there are no carry-overs

for example, take 72133641

its flip pairs are 7&1, 2&4, 1&6, and 3&3. All of those add up to single digits, so there are no carry-overs, hence a perfect palindrome after round 1

note you can take this number in two ways:

716

716 + 617 = 1333

or 0716 + 6170 = 6886

This can obviously be done to any number to manipulate it so that its flip-pairs all add up to single digits... so bonanova, what's your rule on zeroes?

btw congrats on post #1000 ;D

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Now their Example for 196:

196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, 1997

Should it not go like this instead?

196 887 1675 7436....

Hi chanakyavg - I've just looked at the wikipedia page and I think you've just misread the page a little. That first list of numbers above is provided as a list of suspected Lychrel numbers (or "candidates") rather than an example of how 196 progresses.

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Hi chanakyavg - I've just looked at the wikipedia page and I think you've just misread the page a little. That first list of numbers above is provided as a list of suspected Lychrel numbers (or "candidates") rather than an example of how 196 progresses.

Thanks, i read that page again after ur post, seems like I did mis-read it.

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bona - congrats on the 1,000th post.

I'm afraid I don't have much to contribute to this puzzle. I do however, have a theory I'd like to contribute to the thread in the hopes that someone else might be able to run with it. I do this while running the risk of giving away the answer to one of my own posts but seeing as how this one is so much more intriguing I think it's worth it.

I worked as an accountant for a little while after college and before I left that career path running and screaming I did pick up a little trick every good accountant uses. Whenever you're reconciling an account and you end up with a difference, the first thing you do is divide the difference by 9. If the answer is divisible by 9, there's a 99% chance the reason you're off is a transposition error. A lot of people (especially those who are good at number puzzles) outside the accounting world may already know of this math fact but I have also learned that if you ever have a coincidence between two numbers there's a 99% chance it's not a coincidence. And seeing as how this puzzle has a lot to do with number transpositions I can't help but think this little math fact might not just be a coincidence.

For example:

86 - 68 = 18/9 = 2

651 - 165 = 486/9 = 54

7824 - 8274 = -450/9 = -50

Notice it doesn't even matter what sort of transposition takes place, as long as all the digits of the original number are still present, the difference will be a multiple of 9.

Now I know the OP has a strict forward to backward transposition and the numbers are added to together instead of subtracted but like I said, I can't help but feel that this coincidence is more than a coincidence. I'd love to know if anyone has any thoughts on this.

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Very cool post. Please don't give the keyboard a rest. I look forward to your next 1000.

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Happy 1000th post and Easter

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Congrats on 1,000 - can't wait to see more of what is in store for the rest of us on your next 1,000.

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Why would you take this offline? The whole point is to discuss it here

my idea:

if the number is made up of digits that in total add up to less than 10 (ie, add up to a single digit), than you get it in the first round:

2+1 = 3, which is a single digit, so:

21 + 12 = 33

4+1+1+2+1 = 9, single digit, so:

41121 + 12114 = 53235

1117 adds up to 10, but it doesnt necessarily mean it WONT be a palindrome on the first round:

1117 + 7111 = 8228

1218 + 8121 = 9339

So that leads to a more general (and obvious) rule, that if the corresponding flip-pairs each add up to a single digit, there is a palindrome in the first round, because there are no carry-overs

for example, take 72133641

its flip pairs are 7&1, 2&4, 1&6, and 3&3. All of those add up to single digits, so there are no carry-overs, hence a perfect palindrome after round 1

note you can take this number in two ways:

716

716 + 617 = 1333

or 0716 + 6170 = 6886

This can obviously be done to any number to manipulate it so that its flip-pairs all add up to single digits... so bonanova, what's your rule on zeroes?

btw congrats on post #1000 ;D

adding to my own post (which was quoted above, also bonanova has yet to answer my question about 0's, which can make ANY number work on the first round), what ALFRED said reminded of one of the properties of the number 9

the difference between a two-digit number and its inverse is 9 times the difference between the two digits making up the two-digit number

for example, 35+18=53, cuz 9*(5-3)=18

It's very easy to see WHY that happens. That specific method only works with 2-digit numbers, though.

And as ALFRED pointed out, in a sort of generalization of what I was just talking about, the difference between a number and its inverse is always a multiple of 9

so if:

abc - cba = m9 (m9 meaning a multiple of 9)

then

abc+cba = abc-cba+cba+cba

abc+cba = m9 + 2(cba)

453+354 = (453-354) + 2(354)

453+354 = 99 + 2(354)

This is all pretty obvious stuff, but I'm just pointing it out for general use

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bona - congrats on the 1,000th post.

I'm afraid I don't have much to contribute to this puzzle. I do however, have a theory I'd like to contribute to the thread in the hopes that someone else might be able to run with it. I do this while running the risk of giving away the answer to one of my own posts but seeing as how this one is so much more intriguing I think it's worth it.

I worked as an accountant for a little while after college and before I left that career path running and screaming I did pick up a little trick every good accountant uses. Whenever you're reconciling an account and you end up with a difference, the first thing you do is divide the difference by 9. If the answer is divisible by 9, there's a 99% chance the reason you're off is a transposition error. A lot of people (especially those who are good at number puzzles) outside the accounting world may already know of this math fact but I have also learned that if you ever have a coincidence between two numbers there's a 99% chance it's not a coincidence. And seeing as how this puzzle has a lot to do with number transpositions I can't help but think this little math fact might not just be a coincidence.

For example:

86 - 68 = 18/9 = 2

651 - 165 = 486/9 = 54

7824 - 8274 = -450/9 = -50

Notice it doesn't even matter what sort of transposition takes place, as long as all the digits of the original number are still present, the difference will be a multiple of 9.

Now I know the OP has a strict forward to backward transposition and the numbers are added to together instead of subtracted but like I said, I can't help but feel that this coincidence is more than a coincidence. I'd love to know if anyone has any thoughts on this.

Interesting idea.

I'm not sure you can go anywhere interesting [with 9's at least] from sums of transposed numbers.

The difference [which is interesting] and the sum differ by twice one of the numbers [which can be anything].

a propos, you might be interested to revisit this puzzle [even tho it's cover is now blown].

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Neat little math bit there - I have nothing intelligent to add, just to say 'nova - GET A LIFE!! J/k - congrats on enriching our minds with all of your insights

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Neat little math bit there - I have nothing intelligent to add, just to say 'nova - GET A LIFE!! J/k - congrats on enriching our minds with all of your insights

Good advice!

Being retired, I could say that I've already HAD a life, but that's not close to being true.

I still have one [can you spell grandchildren?] and the fun of this site is part of it.

Thanks ... !

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Yikes - well having grandchildren is far harder of a job than I have (guess it's like raising kids a SECOND time, so at least you have some experience). Plus I guess being on here keeps you out of trouble - keep up the good posts, though - I enjoy reading your stuff

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Interesting idea.

I'm not sure you can go anywhere interesting [with 9's at least] from sums of transposed numbers.

The difference [which is interesting] and the sum differ by twice one of the numbers [which can be anything].

a propos, you might be interested to revisit this puzzle [even tho it's cover is now blown].

That is an interesting post. And unreality I would like to hear some more of those magic-9 tricks. I still feel like there is some sort of connection buried in there.

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Well adding 9 is like incrementing the tens place by 1 and decrementing the ones place by 1

17 + 9 =

26

119 + 9 =

128

etc

so adding a multiple of 9 is like doing that multiple times

adding 27 (9*3) is like incrementing the tens place by 3 and decrementing the ones place by 3

example:

55 + 27 =

82

I'm still trying to see how this relates to this riddle, but I think there's probably a connection

I did sorta find one in the other post:

abc+cba = m9 + 2(cba)

where m9 is any multiple of 9, abc is a number (of any amount of digits, i made it abc arbritrarily) and cba is its inverse

on a different point, what IS inversing?

abc would be a*b*c, so what I mean by abc must be:

100a+10b+c

flipped is:

100c+10b+a

so in a three-digit number, flipping means adding 99c and subtracting 99a

is there some sort of pattern we can get for x amount of digits?

a = a, no change needed, keep a the same

10a+b = 10b+a, you need to add 9b and subtract 9a

100a+10b+c = 100c+10b+a, you need to add 99c and subtract 99a, keep be the same

1000a+100b+10c+d = 1000d+100c+10b+a, you need to add 999d, add 90c, subtract 90b and subtract 999a

10000a+1000b+100c+10d+e = 10000e+1000d+100c+10b+a, you need to add 9999e, add 990d, keep c the same, subtract 990b, subtract 9999a

yeah I see a pattern emerging....

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Well adding 9 is like incrementing the tens place by 1 and decrementing the ones place by 1

17 + 9 =

26

119 + 9 =

128

etc

so adding a multiple of 9 is like doing that multiple times

adding 27 (9*3) is like incrementing the tens place by 3 and decrementing the ones place by 3

example:

55 + 27 =

82

I'm still trying to see how this relates to this riddle, but I think there's probably a connection

I did sorta find one in the other post:

abc+cba = m9 + 2(cba)

where m9 is any multiple of 9, abc is a number (of any amount of digits, i made it abc arbritrarily) and cba is its inverse

on a different point, what IS inversing?

abc would be a*b*c, so what I mean by abc must be:

100a+10b+c

flipped is:

100c+10b+a

so in a three-digit number, flipping means adding 99c and subtracting 99a

is there some sort of pattern we can get for x amount of digits?

a = a, no change needed, keep a the same

10a+b = 10b+a, you need to add 9b and subtract 9a

100a+10b+c = 100c+10b+a, you need to add 99c and subtract 99a, keep be the same

1000a+100b+10c+d = 1000d+100c+10b+a, you need to add 999d, add 90c, subtract 90b and subtract 999a

10000a+1000b+100c+10d+e = 10000e+1000d+100c+10b+a, you need to add 9999e, add 990d, keep c the same, subtract 990b, subtract 9999a

yeah I see a pattern emerging....

I agree unreality. You're definitely onto something here. I'm going to play with those formulas and see if I can come up with something.

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I made a word document but now it says I cant upload it, and it has a bunch of superscript and subscript stuff and I dont want to convert to bbcode

Is there any way I can get that file up here, cuz it followed thru with those formulas I just started... it's a Word document

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bona and unreality - I don't know if you've seen this yet but it looks like we've just stumbled upon the tip of the iceberg. This guy has been running computer programs for over a decade to compute the largest iterations of palindromic numbers.

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I made a word document but now it says I cant upload it, and it has a bunch of superscript and subscript stuff and I dont want to convert to bbcode

Is there any way I can get that file up here, cuz it followed thru with those formulas I just started... it's a Word document

1. send me the file and I will upload it if it's ok

2. make a picture of it - decrease size - upload it

btw, subscript and superscript works in this forum

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After I read this thread, the only thing that comes to mind is

"Why do Native American Raindances work?"

They don't stop until it rains

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After I read this thread, the only thing that comes to mind is

"Why do Native American Raindances work?"

They don't stop until it rains

I hate to say it niner but I'm starting to think that's the correct answer.

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