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### #11

Posted 24 March 2008 - 02:15 AM

### #12

Posted 24 March 2008 - 02:23 AM

### #13

Posted 24 March 2008 - 04:25 AM

### #14

Posted 24 March 2008 - 05:00 AM

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

### #15

Posted 24 March 2008 - 06:24 AM

Interesting idea.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.

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].

*Vidi vici veni.*

### #16

Posted 24 March 2008 - 06:32 AM

### #17

Posted 24 March 2008 - 06:39 AM

Good advice!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

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 ... !

*Vidi vici veni.*

### #18

Posted 24 March 2008 - 06:43 AM

### #19

Posted 24 March 2008 - 02:55 PM

Interesting idea.

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

The difference [whichisinteresting] 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.

### #20

Posted 24 March 2008 - 04:08 PM

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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