Is this challenge related to the normal use of that abacus?
E.g. is this a base 10 abacus? Or do you expect people
to "redefine" the meaning of the beads?
Eniac
thats too big, so your method of using the abacus is too complicated.
Its been suggested that, if you use the abacus in a very complicated way, you could represent very very large numbers. The answer to this problem doesn't require any terribly complicated method of using, however its not exactly the easiest method either.
Its been suggested that, if you use the abacus in a very complicated way, you could represent very very large numbers. The answer to this problem doesn't require any terribly complicated method of using, however its not exactly the easiest method either.
But the question is
You can use the Chinese abacus with base 16, the two upper beads stand for 5 each, the lower 5 represent 1 each, so 5+5+5=16-1. This one could be used with the left five as the ones and the right five as 6 each, so 5*6+5*1=36-1, then you can represent 36^10-1 as the biggest number. Complicated, true, but without destroying or rearranging.What is the biggest number that can be displayed using this abacus (without destroying or rearranging it)?
Ugh I could represent 2^91 -1 with that... Rather big number I'd say... roughly 27 digits in decimal. But it won't take it 
If I use the beads at bits, where (as with a Compact Disk) two beads touching represents a 0 and a space represents a 1 (or you could take chance vs. not change or a different interpretation...), you'd have 9 binary digits per row. There being 10 rows, you'd have 9*10 digits which makes the highest number the one where all digits are 1's, 2^91 -1.
But it doesn't like my idea or something
If I use the beads at bits, where (as with a Compact Disk) two beads touching represents a 0 and a space represents a 1 (or you could take chance vs. not change or a different interpretation...), you'd have 9 binary digits per row. There being 10 rows, you'd have 9*10 digits which makes the highest number the one where all digits are 1's, 2^91 -1.
But it doesn't like my idea or something
Ok, here is my calculation, I believe it's mathematically accurate, and that it represents the biggest possible number. Please let me know if there is anything wrong with my reasoning:
1) Every row is a 15 slot, base 3 string, which gives you 3^15 possibilities. Each row starts with the string AAAAA-AAAAA-AAAAA (where 'A' represents an empty slot), and keeps increasing until it reaches CCCCC-CCCCC-CCCCC.The default setup when you load the challenge is AAAAA-BBBBB-CCCCC
2) Since you need to count only the combinations where you start (from the right) with 5 Cs followed by 5 Bs (with As allowed anywhere in between) You have a total of 3003 possible combination for each row, this can be verified programmatically by iterating over the 3^15 combinations and counting the ones that follow this pattern.
3) Since you have 10 of these rows, the biggest number should be 3003^10 which equals 59642154303295182755378537795549049
1) Every row is a 15 slot, base 3 string, which gives you 3^15 possibilities. Each row starts with the string AAAAA-AAAAA-AAAAA (where 'A' represents an empty slot), and keeps increasing until it reaches CCCCC-CCCCC-CCCCC.The default setup when you load the challenge is AAAAA-BBBBB-CCCCC
2) Since you need to count only the combinations where you start (from the right) with 5 Cs followed by 5 Bs (with As allowed anywhere in between) You have a total of 3003 possible combination for each row, this can be verified programmatically by iterating over the 3^15 combinations and counting the ones that follow this pattern.
3) Since you have 10 of these rows, the biggest number should be 3003^10 which equals 59642154303295182755378537795549049
You like pink, don't you?
-
polarlemniscate
- Posts: 6
- Joined: Wed Mar 03, 2010 1:56 pm
ENIAC
Think about number bases - it's really very simple. I suspect the ENIAC title is something to do with this. What number base did ENIAC work in?