question for Roman math geeks

Discussion in 'Ancient Coins' started by rrdenarius, Jun 4, 2019.

  1. rrdenarius

    rrdenarius non omnibus dormio Supporter

    Can you solve this equation? 20190604_155535.jpg
     
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  3. Aleph

    Aleph Well-Known Member

    S=VI?
     
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  4. kaparthy

    kaparthy Well-Known Member

    Five equals six minus one.
     
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  5. 7Calbrey

    7Calbrey Well-Known Member

    5= 4+1(squarre)
     
  6. Alegandron

    Alegandron "ΤΩΙ ΚΡΑΤΙΣΤΩΙ..." ΜΕΓΑΣ ΑΛΕΞΑΝΔΡΟΣ, June 323 BCE

    Isnt ‘V’ a Quinarius (5 Asses), an ‘S’ a Semis (Half-Ass), and I’s are an As.

    so... my dyslexic logic feels that 2 RR Sestertii (or SII as a denomination - which means 2Asses and a Semis or 2-1/2 Asses) = a V or a Roman Quinarius.
     
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  7. rrdenarius

    rrdenarius non omnibus dormio Supporter

    I was going for V is a Roman 5. S is six ounces. The square root of -1 is i, so i squared is -1.
     
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  8. ancientcoinguru

    ancientcoinguru Supporter! Supporter

    I thought 5=6-1 was too obvious, so I was trying to work this out like @Alegandron (I enjoyed your solution Brian:joyful:)
     
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  9. kaparthy

    kaparthy Well-Known Member

    It is not a 1, it is an i the symbol for the square root of -1.

    But I would with the sesterius and asses, but it does not add up.
     
  10. Alegandron

    Alegandron "ΤΩΙ ΚΡΑΤΙΣΤΩΙ..." ΜΕΓΑΣ ΑΛΕΞΑΝΔΡΟΣ, June 323 BCE

    LOL, like every thing they did in the Roman Empire, it DOES add up when they want to MAKE it correct for their ends. :D

    It only takes creativity... and I DID explain that I truly am dyslexic.

    Thank you @ancientcoinguru
     
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