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The Abacus Challenge

MathCookieMathCookie Member Posts: 286 ✭✭
edited July 2017 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


  • idcidc Member Posts: 139 ✭✭✭
    edited July 2017
    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.

    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: 286 ✭✭
    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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