# Ultimate number trick 2 (no recs needed)

MooTheCow
Posers and Puzzles 07 Jan '05 03:02
1. 07 Jan '05 03:02
Here's an interesting trick that hasn't been mathematically proven, but works. I thought of it last week.

Steps (example number in parentheses):

1. Think of any whole number greater than 2 (5)
2. Subtract 1 (4)
3. Square the number that you get in the previous step (16)
4. Multiply by 4 (64)
5. Add 1 at the end of the number in the previous step (641)
6. Add the number in step 5 to itself backward (641+146=787)
7. Repeat step 6 as often as necessary, and you'll eventually get a palindrome (you do not need to do this step that much, it normally produces a palindrome quite quickly)

Note: For those of you who don't know, a palindrome is a number that's the same backward and forward (1221, 141, 3548453, 22, 38983, etc)

So, what's so special about this? Well, if you try to put a very unlucky number in step 1, you'll never, never, never get a palindrome no matter how much you repeat step 6.

Hint: That number is between 10 and 20.
2. Acolyte
Now With Added BA
07 Jan '05 21:311 edit
Originally posted by MooTheCow
Here's an interesting trick that hasn't been mathematically proven, but works. I thought of it last week.

Steps (example number in parentheses):

1. Think of any whole number greater than 2 (5)
2. Subtract 1 (4)
3. Square the num ...[text shortened]... much you repeat step 6.

Hint: That number is between 10 and 20.
Well, there's nothing interesting about 1-5, as far as I can tell. The interesting thing is 6. and 7., where you repeatedly reverse and add until you get a palindrome. We'll call a number unlucky if this never happens. (Forget about 1-5, so we'll say your example starts with 641.)

The interesting question now is: which numbers are unlucky? Your claim is that you've found such a number, but how would one prove that it is indeed unlucky?

One thing I can say fairly quickly: all numbers below 55 are lucky. It's clear why this is the case for one-digit numbers, and for a two-digit number, the first iteration will give us a multiple of 11, and all of those below 154 give a palindrome in at most one step. I'll have to think about the general case, but the following might help: when adding the numbers, consider the sums of the individual digits before carrying 1s etc.
3. 08 Jan '05 05:01
The unlucky number is 13
4. 08 Jan '05 05:031 edit
Steps (example number in parentheses):

1. Think of any whole number greater than 2 (13)
2. Subtract 1 (12)
3. Square the number that you get in the previous step (144)
4. Multiply by 4 (576)
5. Add 1 at the end of the number in the previous step (5761)
6. Add the nu ...[text shortened]... as often as necessary, and you'll eventually get a palindrome (never)
Steps (example number in parentheses):

1. Think of any whole number greater than 2 (13)
2. Subtract 1 (12)
3. Square the number that you get in the previous step (144)
4. Multiply by 4 (576)
5. Add 1 at the end of the number in the previous step (5761)
6. Add the number in step 5 to itself backward (5761+1675=7436)
7. Repeat step 6 as often as necessary, and you'll eventually get a palindrome (you do not need to do this step that much, it normally produces a palindrome quite quickly
5. BigDoggProblem
08 Jan '05 08:40
Originally posted by MooTheCow
Steps (example number in parentheses):

1. Think of any whole number greater than 2 (13)
2. Subtract 1 (12)
3. Square the number that you get in the previous step (144)
4. Multiply by 4 (576)
5. Add 1 at the end of the number in the previous step (5761)
6. Add the number in step 5 to itself backward (5761+1675=7436)
7. Repeat step 6 as often ...[text shortened]... rome (you do not need to do this step that much, it normally produces a palindrome quite quickly
Actually, the real unlucky number is 5761. The first steps are just to allow 13 to be a 'solution'. It is true that your first 5 steps drastically reduce the number of solutions.

I am curious why you would say that 5761+1675 etc. never leads to a palindrome. How would you know that it never does?

I wrote a program to solve this problem. It repeats step 6 thirty times for each number from 2 up to a number I provide. It is not an absolute proof, but it is still a good test.

According to my program, 13 is not the only answer to this problem! 60 also works:
1. Think of any whole number greater than 2 (60)
2. Subtract 1 (59)
3. Square the number that you get in the previous step (3481)
4. Multiply by 4 (13924)
5. Add 1 at the end of the number in the previous step (139241)
6. Add the number in step 5 to itself backward (282172)
And repeating Step 6 gives:
Result 1 = 282172
Result 2 = 553454
Result 3 = 1007809
Result 4 = 10094810
Result 5 = 11943811
etc. with no palindrome in sight.

Other solutions found are 80, 99, 110, etc.
6. 09 Jan '05 05:23
I know that it plugging in the number 13 will never produce a palindrome because I tried repeating step 6 200 times earlier, but nothing works.

Before your post, I did not know that the number 60 was also unlucky. Anyway, do the unlucky numbers occur more often when the number in the first step is high (&gt;1000)
7. 10 Jan '05 07:491 edit
Originally posted by MooTheCow
I know that it plugging in the number 13 will never produce a palindrome because I tried repeating step 6 200 times earlier, but nothing works.

Before your post, I did not know that the number 60 was also unlucky. Anyway, do the unluck ...[text shortened]... occur more often when the number in the first step is high (>1000)
For this &quot; ultimate number trick-2&quot;, it seems, there are more exceptions than adherents, as you go for higher and higher numbers. BTW- step 1 in the trick is unnecessary, Step 2 can straightaway be step 1 with the number to be selected being more than 1, and the later steps remaining as it is..