The New Sudoku Players' Forum

[Cec]

You may know this one but it is intriguing.

a. Write down any three numbers (eg 689)
b. Reverse these numbers (986)
c. Calculate the difference (just simple subtraction)
d. Reverse these three numbers
e. Add your answers together (c+d)

The first two replies will reveal the mystery for those who don't already know it.

Cec

[Ruud]

Now turn the answer upside down.

What am I?

[Cec]

Now turn the answer upside down.
What am I?

Hey! what about you answering first
I've turned my answer upside down and it's just four different numbers. I don't get it ?
Cec

[vidarino]

Now turn the answer upside down.

What am I?

You're a microprocessor?

Anyway, stepping back in line, my answer is 1089.

[Ruud]

I've turned my answer upside down and it's just four different numbers. I don't get it ?
Cec

I got the same answer each time. Just needed the variation

The answer could have been January '68.

Ruud.

[Cec]

"..Anyway, stepping back in line, my answer is 1089. "

'1089' eh! Thanks vidarino. The next answer posted will reveal the mystery or perhaps it's revealed already? On Ruud's above "turning the answer upside down" I'm still in the dark on this. Any more clues Ruud? - when you post your answer that is.
Cec
PS. Noticed Ruud has posted in before me but I'll let this stand.
Cec

[Ruud]

OK, let's try to analyze this.

there are 2 possible answers. 1089 and 0.

Let's take a number constructed of digits A, B & C.

B is irrelevant. When you reverse the number, B remains in the same spot, and then it is subtracted from B.

When A equals C, both A and C become irrelevant. The answer will always be 0. Step d. Does not work in that case, because you do not have 3 numbers. So maybe the question should have been:

a. Write down any three different numbers.

What happens when A != C?

Assume A > C. You can always swap the 2 digits to make A > C.

Then we have a result in c. (XYZ) where X = A-C-1, Y=10-1, Z = 10+C-A.

Reverse and add the 2 together makes you lose all A and C variables:

9 * 100 + 180 + 9 = 1089 (which, by the way, is 6801 upside-down, but that too is irrelevant)

You can also continue:

Divide the number by 9.
Take the square root.
Now think of a month.

Were you thinking of the month November?

Ruud.

[Cec]

"OK, let's try to analyze this..."

Most intriguing - I'm lost for words to express my thoughts on your wisdom and participation.
Cec

[Cec]

"OK, let's try to analyze this..."

Most intriguing - I'm lost for words to express my thoughts on your wisdom and participation.
Cec

After enduring flu symptoms yesterday when everything seemed annoying I am able today to much better appreciate Ruud's above explanation.

I've long realized there had to be a mathematical reason why any of the three chosen numbers in this exercise would always give the same answer (1089 - or zero if the first and third numbers are identical as pointed out by Ruud) but haven't read or heard of this logical explanation until now.

I also liked Ruud's extension of the exercise to arrive at the 11th month as the answer. Thanks to Ruud and Vidarino for participating.
Cec

[udosuk]

Cec, if you like these stuffs, let's try this:

1. Pick any 4-digit number, as long as it's not 4 identical digits, such as 1111 or 5555.

2. Rearrange the digits in descending order, e.g. 1795 -> 9751, 4096 -> 9640.

3. Reverse the previous result to get another number, e.g. 9751 -> 1579, 9640 -> 469.

4. Calculate the difference between the previous 2 results, e.g. 9751-1579=8172, 9640-469=9171. If the difference has less than 4 digits, pad some zeros in front and pretend it to be a "4-digit number", e.g. 27 -> 0027, 343 -> 0343.

5. Repeat steps 2 to 4, e.g. 8172 -> 8721-1278=7443, 9171 -> 9711-1179=8532. Do this for 7 times or until you get the same result as the last cycle (then there is no point in recycling).

What's your final result?

[MCC]

I started with:
8725 which gives
8752-2578=6174

6174
7641-1467=6174

Result = 6174.

Is that correct

[Cec]

Cec, if you like these stuffs, let's try this..
..What's your final result?

Actually a headache which wasn't your fault. Fortunately MCC's above reply enlightened me only after I posted a lot of waffle in trying to understand how to get a result. I deleted that post hopefully before too many read it.

Look's like MCC is correct.

My first random selection was:
2684 --> 8642 - 2468 = 6174 (same as MCC)

I then tried:
3761--> 7631 - 1367 = 6264 so repeat process
6264--> 6642 - 2466 = 4176 ......ditto
4176--> 7641 - 1467 = 6174 ......ditto
6174--> 7641 - 1467 = 6174... ' Bingo' same answer

Very interesting - thanks udosuk
Cec

[Ruud]

You are right, Cec. This problem is a little more complicated.

Every combination of 4 digits leads to the answer 6174.

The problem is chaos-related. 6174 is a "strange attractor" in the iterative process. (steps 2-4)

The attractor is reached somewhere between 1 and 8 steps.

I plotted a chart for all digit combinations, with y being the first 2 digits and x the last 2. Each color represents a number of iterations (blue=1, red=8)



No easy reasoning that explains the convergeance.

But these chaos problems usually produce nice pictures

Ruud.

[Cec]

You are right, Cec. This problem is a little more complicated.

An interesting study Ruud but I'm feeling humbled in being seen as "right" because there is a funny side to all this. It was actually MCC who unknowingly steered me in the right direction as to how to methodically follow udosuk's correctly defined procedures which my mind couldn't properly absorb - I could use a nagging Flu bug as my defence...AhhChew!. pardon me!

I was confused why udosuk referred to "rearranging the digits in descending order, e.g. 1795 -> 9751 which I thought should have read an "ascending" order since the "number" 1795 had increased to 9751.
Most of my two to three hours was spent on drafting a post in trying to explain my confusion and not being able to work the procedures out. As I said, fortunately MCC saved me without knowing it but to my relief enabled my initial senseless post to be deleted - I hope all this makes sense.
Cec

[udosuk]

Guys, glad you like this.

If you're interested to read my analysis on this mystery, here is a thread I started in another forum... 5 posts all by myself...

Once again I'm impressed by Ruud's graphical approach on this...

And Cec, no worries about anything, I haven't read your previous reply when you haven't understood the procedure correctly... And believe me I've on that side too... Glad you sorted it out later... Get well soon!