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In calculating the number of Mikvas that the "Sea of Shlomo" could hold, the Gemara (Eruvin 14b) calculates that it would hold 150. The Gemara asks, that being that the Sea was round, and the area/volume of something round is one fourth that of something square of the same diamater, the Sea should have held 125, not 150 Mikvas. Rashi (s''v Michdi) explains that of the 500 cubic Amot in a block of 10 by 10 by 5 Amot, take away a fourth from the first 400, making 300. And a Mikva is 3 cubic Amot, therefore we have 100 Mikva'ot. Of the last 100 Amot, there are 25 Mikva'ot, "as each Mikva is 4 Amot, and these four are really three, as they are short Amot."

Two questions: Would it not be simpler to say "subtract a fourth from 100 and you have 75, and each Mikva is 3 cubic Amot, so there are 25 Mikva'ot"? Also, what does Rashi mean when he says that "these four are really three, but are short Amot"?

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Just a comment on the simpler math: We have much better mathematical tools these days. Just look at how hard the Rambam works to avoid division when calculating the new moon. –  Ariel Sep 4 '12 at 0:05
    
There are two measures of אמות — a woman's אמות and a man's אמות. (I will look for the source on that.) –  b a Sep 4 '12 at 0:05
    
Related: judaism.stackexchange.com/q/16883. –  msh210 Sep 4 '12 at 3:32
    
@Ariel, exactly. Also in Pesachim where it's calculating the volume reviis halog (minimum size of cup of wine) in terms of cubic "fingerwidths" -- the Rishonim do the math in all sorts of convoluted orders to avoid fractions! –  Shalom Sep 4 '12 at 9:13
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