Think about it

1 = 5
2 = 505
3 = 5005
4 = 50005
5 =

?

Originally posted by wolfgang59
1 = 5
2 = 505
3 = 5005
4 = 50005
5 =

?

5 = 1

as 1 = 5

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

Originally posted by coquette
5 = 1

as 1 = 5

correct

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

Your solution is not unique. 55000, 50500, 50050 also fits your requirements.

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.

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.

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!