#### Posers and Puzzles

wolfgang59
Posers and Puzzles 30 Dec '13 01:48
1. wolfgang59
30 Dec '13 01:48
1 = 5
2 = 505
3 = 5005
4 = 50005
5 =

?
2. coquette
30 Dec '13 07:15
Originally posted by wolfgang59
1 = 5
2 = 505
3 = 5005
4 = 50005
5 =

?
5 = 1

as 1 = 5
3. 02 Jan '14 16:59
the digits of each number can be rearranged to give n other valid numbers (i.e. without leading zeros)

e.g.
505, 550
5005,550,5050
4. wolfgang59
02 Jan '14 22:24
Originally posted by coquette
5 = 1

as 1 = 5
correct
5. 03 Jan '14 22:33
Originally posted by wolfgang59
1 = 5
2 = 505
3 = 5005
4 = 50005
5 =

?
but, it is ambiguous:

1 = 5 which has 1 possible arrangement (5)
2 = 505 which has exactly 2 possible arrangements (505, 550)
3 = 5005 : 3 arrangements (5005, 5050, 5500)
4 = 50005: 4 arrangements (50005, 50050, 50500, 55000)

so the fomula could be:
for any number n, the lowest number, made of only the digits 0 and 5, which can be rearranged into n distinct legal numbers.

so
5 = 500005
6. 03 Jan '14 23:21
Your solution is not unique. 55000, 50500, 50050 also fits your requirements.
7. 04 Jan '14 10:091 edit
Originally posted by Sebastian Yap
Your solution is not unique. 55000, 50500, 50050 also fits your requirements.
As I said:

for any number n, the lowest number, made of only the digits 0 and 5, which can be rearranged into n distinct legal numbers, i.e. 50005. This also fits the given numbers.
8. talzamir
Art, not a Toil
18 Jan '14 23:28
f(n) = 5 + 5 x 10^n would be good for most cases but not f(1).. oh well.

"lowest POSITIVE number.." and Tiger's wording is even more flawless, eliminating negative numbers and 1 = 0.
9. 20 Jan '14 09:00
Originally posted by talzamir
f(n) = 5 + 5 x 10^n would be good for most cases but not f(1).. oh well.

"lowest POSITIVE number.." and Tiger's wording is even more flawless, eliminating negative numbers and 1 = 0.
Good point talz!