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The Off Topic section is not meant for discussing Cookie Clicker.

The Abacus Challenge

MathCookieMathCookie Member Posts: 260 ✭✭
edited July 23 in Off Topic
How high can you count using this abacus?
Explain your system on how high you can count using the Abacus.
Here are the rules:
1. Every positive integer up to the largest number representable by the system must be representable in a unique way. So if the system can go up to 10 million, the arrangement of beads for 2 can't be the same as the arrangement for 3,429,501, for example.
2. No modifications can be done to the abacus. For example, you can't add an 11th bead to a row.
3. No using things other than the abacus. So no rulers, for an example.

I can't really think of any other rules that aren't ones that don't need to be said.

This competition is like my large number competition, in that you want to name a system with a higher upper limit than the last person's system.

I'll start off way too easy, with a system where you just move each bead over from left to right to count, and once you complete a row, do the same to the next row. (100)
Post edited by MathCookie on

Comments

  • idcidc Member Posts: 130 ✭✭✭
    edited July 23
    There are 100 total beads.
    If you space them all out evenly you can then move each bead to the left a bit for 1, keep it where it is for 0 and to the right a bit for a 2. You now can represent numbers up to 3^100 ≈ 5.154*10^47

    You could do larger numbers than 3 but it wont be as easy to see the number.

    There should be a rule that you can't use a ruler. Otherwise you get into a situation where a hypothetical ruler can measure the Planck length.

    Edit:
    If it is too hard to see which are shifted, you could keep the first row to check against. You can still do numbers up to 3^90 ≈ 8.728*10^42
    BEETLE there is! While the ROOMS here not!
  • MathCookieMathCookie Member Posts: 260 ✭✭
    Wow. I had an idea for a binary system that went up to 2^100 but I just got rekt

    Here's an extension on yours: Keeping where it is is 0, Left a bit is 1, left more (like 1/2 of the way to the previous bead's 0 position instead of a fourth) is 2, right a bit is 3, right More is 4.

    5^100 ≈ 7.888*10^69.

    I'd say that the difference between 1 and 2 in my system is easy to see on an actual abacus, but going past that would be hard to see, I agree. (I said 1/2 of the way because it can't go farther- if bead 1 had a 4 and bead 2 had a 2, they'd be touching, soooo...)

    (And this thread's probably gonna die quickly lol)
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