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This answer ought to be much higher - e.g. a single bead on a row can be in one of three states (left, middle, right), two beads have seven states I think (both left, both together in the middle, both right, and four variations with a gap between the two beads), ...

gfoot wrote:This answer ought to be much higher - e.g. a single bead on a row can be in one of three states (left, middle, right), two beads have seven states I think (both left, both together in the middle, both right, and four variations with a gap between the two beads), ...

...resulting in (2^11-1)^10 = 1291749870339606615892191271170049 distinguishable states for the shown abacus. However, it is very fiddly to use it in that way.

Anyway, what is the connection between this challenge and the old Eniac?

It would be eaiser, if you know how real abacus in China works

Anyway, what is the connection between this challenge and the old Eniac?

This is just a joke, as the abacus shown in this challenge is almost as "new" and almost as powerful as good old Eniac But since many people are misleaded by this name it serves another purpose, too

This answer ought to be much higher - e.g. a single bead on a row can be in one of three states (left, middle, right), two beads have seven states I think (both left, both together in the middle, both right, and four variations with a gap between the two beads), ...

This would lead much too far - if the distance between the single beads would matter, there would be infinite possibilities or states. Here, a bead can simply signify/have two states: on and off.

Seriously, why is the accepted solution 11^10 -1 = 25937424600?

The way you use an abacus with 10 beads per row properly, which is described here:
http://www.differ.freeuk.com/learning/m ... works.html

you can push at most 9 beads (out of 10) to the one side of the respective row. If you add +1 in that row, the row gets "reset" and the next higher one gets +1. Thus, the highest number that can be shown on a 10x10 abacus:
= 9 999 999 999

It can't be higher than that, because if you move any more bead to the left side, it causes an "overflow" up until the highest column, which cannot represent more than what it does in this example.

AMindForeverVoyaging wrote:Seriously, why is the accepted solution 11^10 -1 = 25937424600?

The way you use an abacus with 10 beads per row properly, which is described here:
http://www.differ.freeuk.com/learning/m ... works.html

you can push at most 9 beads (out of 10) to the one side of the respective row. If you add +1 in that row, the row gets "reset" and the next higher one gets +1. Thus, the highest number that can be shown on a 10x10 abacus:

Code: Select all

o-ooooooooo  10^9 * 9 = 9 * 1 000 000 000
o-ooooooooo  10^8 * 9 = 9 *   100 000 000
o-ooooooooo  10^7 * 9 = 9 *    10 000 000
o-ooooooooo  10^6 * 9 = 9 *     1 000 000
o-ooooooooo  10^5 * 9 = 9 *       100 000
o-ooooooooo  10^4 * 9 = 9 *        10 000
o-ooooooooo  10^3 * 9 = 9 *         1 000
o-ooooooooo  10^2 * 9 = 9 *           100
o-ooooooooo  10^1 * 9 = 9 *            10
o-ooooooooo  10^0 * 9 = 9 *             1

= 9 999 999 999

It can't be higher than that, because if you move any more bead to the left side, it causes an "overflow" up until the highest column, which cannot represent more than what it does in this example.

Considering that a bead can assume two and only two different states (to the left or to the right), you have eleven possible states per row (all to the left, 9 states where beads are divided between left and right plus all beads to the right), thus 11^10 possibilities for 10 rows. It is simply an eleven-based arithmetics.

Except base 11 does not make any sense. Abacus (Abaci?) with ten beads per row do exist, and nobody uses them for base-11 arithmetic, because that's just useless.

If you want some weird base that nobody in real life uses, then you need to show a picture of a "constructed" abacus with, say, 11x7 beads or some such. Not a picture of an abacus that actually exists, and is absolutely and without any doubt used for base-10 calculation.

So you really disagree that there are 11 different states for each row? The challenge does not say that the abacus should be used in a 'traditional' way. The important thing is that it can be used to do base-11 calculation.

The thing is, if you can use the abacus in a non-traditional way, then even higher maximum numbers are possible, as gfoot has pointed out.

So, no matter how you put it, the accepted answer is wrong. It is either too high (if you use the abacus in a traditional way), or too low (if you can use it arbitrarily, without breaking it as stated in the challenge).

bsguedes wrote:The important thing is that it can be used to do base-11 calculation.

I am able to use a key to open a bottle of beer. Does that mean that this is the expected way of using a key?

AMindForeverVoyaging wrote: I am able to use a key to open a bottle of beer. Does that mean that this is the expected way of using a key?

You can, but the appropriate tool is a chainsaw, as we all know.