Clearly the true value of pi is really 3.174, and this constant is just (pi +3)×1000.
6174 = Golden Bay, Western Australia.
party_pants said:
6174 = Golden Bay, Western Australia.
Getting odder and odder.
Interesting. WP gives 495 has the 3 digit equivalent.
There appears to be no corresponding 2 digit equivalent: performing this algorithm just results in cyclical behaviour.
In base 10, it can easily be shown that all numbers of the form 6174, 631764, 63317664, 6…333…17…666…4 (where the length of the “3” sequence and the length of the “6” sequence are the same) are fixed points of the Kaprekar mapping.
The Rev Dodgson said:
Clearly the true value of pi is really 3.174, and this constant is just (pi +3)×1000.
No, pi is unarguably 4. :D
Spiny Norman said:
The Rev Dodgson said:
Clearly the true value of pi is really 3.174, and this constant is just (pi +3)×1000.
No, pi is unarguably 4. :D
Good point.
The Rev Dodgson said:
Spiny Norman said:
The Rev Dodgson said:
Clearly the true value of pi is really 3.174, and this constant is just (pi +3)×1000.
No, pi is unarguably 4. :D
Good point.
Although I don’t know where the pi = 4! comes from.
The Rev Dodgson said:
The Rev Dodgson said:
Spiny Norman said:No, pi is unarguably 4. :D
Good point.
Although I don’t know where the pi = 4! comes from.
Manhattan
This works if the perimeter is infinity I guess
wookiemeister said:
This works if the perimeter is infinity I guess
It works because it never becomes a circle.
sibeen said:
wookiemeister said:
This works if the perimeter is infinity I guess
It works because it never becomes a circle.
You could go the other way perhaps ?
Dunno
No pencil
wookiemeister said:
You could go the other way perhaps ?Dunno
No pencil
Nup. expand a circle and you get a circle.
transition said:
Spiny Norman said:
How delightfully odd
watched that, quite good
OK, I get it.
Looks like 6174 is the only four digit number (other than all digits equal) that satisfies the recurrence relation: digits in descending order minus digits in ascending order equals original number. And many other 4 digit numbers are attracted to it.
The attraction is quite strong. The many to one can be seen from: 9876 – 6789 = 4321 – 1234 = 3087. That means that in this case 144 different initial numbers are attracted to the answer 3087 as the second step. It doesn’t take too many powers of 144 to exceed 9999.
8730 – 0378 = 7992. 9972 – 2779 = 7173. 7731 – 1377 = 6354, 6543 – 3456 = 3087.
Hey, it doesn’t work. We’ve got a loop here, not an attraction to 6174.
mollwollfumble said:
transition said:
Spiny Norman said:
How delightfully odd
watched that, quite good
OK, I get it.
Looks like 6174 is the only four digit number (other than all digits equal) that satisfies the recurrence relation: digits in descending order minus digits in ascending order equals original number. And many other 4 digit numbers are attracted to it.
The attraction is quite strong. The many to one can be seen from: 9876 – 6789 = 4321 – 1234 = 3087. That means that in this case 144 different initial numbers are attracted to the answer 3087 as the second step. It doesn’t take too many powers of 144 to exceed 9999.
8730 – 0378 = 7992. 9972 – 2779 = 7173. 7731 – 1377 = 6354, 6543 – 3456 = 3087.
Hey, it doesn’t work. We’ve got a loop here, not an attraction to 6174.
9972 – 2779 ≠ 7173, although the correct 9972 – 2799 = 7173. Also, 8730 – 0378 = 8532; 8532 – 2358 = 6174.
