
Will the total number of digits in the positions in this market be evenly divisible by the total number of participants?
30
580Ṁ4116resolved Jul 2
Resolved
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At the close of this market, I will look at each position in the market, take the number of digits in it (so, a $1,000 mana position is four digits) and add up all the digit counts.
If that number is divisible evenly, without remainder, by the number of participants - the market resolves YES. Otherwise NO. I will place an initial stake in the market because everything divides by one and no one wants a divide by zero error!
Example:
A has Ṁ 327,789 in YES, B has Ṁ 2,476 in YES, C has Ṁ 5 in NO.
6 + 4 + 1 = 11, 11 / 3 = 3 remainder 2, market resolves NO.
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