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APIS Translation (English)

(Frg. I, col. i);1/23 = 1/23;[2/23] = 1/12 + 1/276;[3/23] = 1/10 + 1/46 + 1/115;[4/23 = 1/6] + 1/138;[5/23 = 1/6 + 1/23] + 1/138;[- - -];;(Frg. I, col. ii);12/29 = 1/4 + 1/8 + 1/29 + 1/232;13/29 = 1/3 + 1/10 + 1/87 + 1/290;14/29 = 1//4 + 1/5 + 1/58 + 1/116 + 1/145;15/29 = 1/2 + 1/58;16/29 = [1/2] + 1/29 + 1/58;[17/29 = 1/2 + 1/12] + 1/348;[- - -];;(Frg. I, col. III: too fragmentary to be translated: table of 1/31?);(Frg. I, col. IV: too fragmentary to be translated);(Frg. II: too fragmentary to be translated);(Frg. III, i: too fragmentary to be translated);;(Frg. III, col. ii);To what number [may you add] 15, subtract 20, [add 25], and subtract 30; it will be 100 artabai of wheat? As is needful, add together the added artabai, = 400; add together also the subtracted artabai, [= 50]; from these subtract the added artabai, [40, =] 10 artabai; add to the 100 artabai of wheat, = 110; this is the number. Proof: Add 15, = 125, [and] subtract 20; remainder, 105. Add [25] = 130; subtract 30; remainder, the given 100 artabai of wheat.;[- - -];;(Frg. III, col. iii);(Reduce to silver drachmai:) 43 talents 2000 (drachmai) of copper, on the basis of one stater to 1200 copper (drachmai). As is needful, reduce the 43 talents 2000 (drachmai) to myriads of drachmai; take 1/1200 of these = 216 2/3 staters; multiply [these] by 4 (since the stater contains 4 drachmai), = [866] drachmai 4 obols.;(Reduce to silver drachmai:) 5800 copper drachmai, on the basis of one stater to 1200 copper drachmai. As is needful, I took 1/1200 of the given 5800 drachmai, = 4 1/2 1/3 staters; multiply these by 4 (since the stater contains 4 drachmai), = 19 drachmai 2 obols.;(Reduce to copper drachmai:) 3 talents 2000 (drachmai) of silver, on the basis of one stater to 1200 copper drachmai. As is needful, reduce the [3 talents 2000 drachmai to] myriads [of drachmai, = 20,000; of these take 1/4,] because the stater [contains 4 drachmai; = 5000] staters; [multiply these into the] 1200 copper [drachmai, = 6,000,000] copper [drachhmai, = 1000 talents of] copper.;[- - -];;(Frg. III, col. iv);Of 100 loads, some are at the rate of 10 drachmai, some at 20 drachmai, and some at 20 drachmai, and some at 30 drachmai; the whole price is 2500 drachmai. [As] is needful I multiplied the 100 loads x 30 drachmai = 3000; from this subtract the given 2500 drachmai; remainder, 500 drachmai. And subtract the 10 drachmai from the 30 drachmai; remainder [20]. I took 1/20 of the remainder, 500 drachmai, = 24 with remainder of 20; and subtract the 20 drachmai from the 30 drachmai; remainder 10 of the original. And one operates with 20 because it is necessary to subtract the 10 drachmai from 30; remainder, 20 once = 20; take besides 20 (since I took 1/20 of the 500 drachmai and it was 24 with remainder of 20), = 40; take 1/10 of this (because you have subtracted 20 drachmai from 30 drachmai and 10 were left), = 4; = 1, since [4 x 1] = 4, for [I took 1/4] of [4]; and subtract 1 from 24 (since I took 1/20 of the [remainder], 500 drachmai, and it was [24 with remainder of 20]); 23 is left; so many are the loads at [10] drachmai, = 230 drachmai. Put above the [10 drachmai] the 23 loads and below them the 230 drachmai. I took [10 from the] 20 drachmai, = [10, and] 1/10 of the 40 drachmai, [= 4; and] put [above the 20 drachmai the] 4 loads, [and under the 20 drachmai the] 80 [drachmai;] add 4 [+ 23 = 27]; subtract [230 drachmai from] 2[500 drachmai, = 2270 drachmai; and subtract 80 drachmai from 22]70 drachmai = 21[90 drachmai; and subtract the 27 loads from the 100 loads = 73; add 23 + 4 + 73 = 100, and] add [270 + 80 + 2190 drachmai = 2500 drachmai];[- - -];;(Frg. III, col. v);Add 1/2 to 2 = 2 1/2; and [add] the [2/3] represented by the 20 stadia to the 3 schoeni = 3 2/3; reduce the 2 1/2 to halves; it is 5; and the 3 2/3 to halves; it is 7 1/3, [of which] 1/5 = 1 1/5 1/6 1/10; (this multiplied) into 4 1/2 artabai = 6 [1/2 1/10 and] (multiplied) into the 60 artabai = 396, 1/100 of which = 3 1/2 1/4 1/5 1/100. So much they pay for 60 artabai over the distance of 3 schoeni 20 stadia.;The freight on [100 artabai] for 3 schoeni is 3 1/2 [artabai]; according to schedule there is a freight charge of 60 artabai; how many artabai will be delivered over a distance of 4 schoeni? As is needful I took [1/3] of the 4 schoeni [=] 1 1/3; (this multiplied) into the 3 1/2 artabai = 4 [2/3, (which reduced) to] thirds = 14; and the 60 artabai (reduced) to thirds = 180, 1/14 [of which] = 12 1/2 1/3 1/42; (this multiplied) into 100 artabai = 1285 2/3 1/21; so many (artabai) will be delivered for 60 artabai [over a distance of] 4 schoeni.;[- - -];;(Frg. III, col. vi);The freight on 100 artabai is [5 artabai]; given a charge of 80 artabai how many artabai in the principal? As is needful, I took [1/5 of] 80 = 16; (multiplied) into 100 = 1600; so many (artabai) in the principal.;The freight on 100 artabai is 5 artabai; what is the freight on the whole principal amount of 1000 artabai? As is needful I multiplied [5] x 1000 = 5000, of which 1/100 = 50; so many (artabai) are the freight.;The width of a field is [2 1/2] schoenia; what is the length that there may be 20 arourai? As is needful, reduce the [2 1/2 schoenia to] halves, = 5; and the 20 arourai to halves, = 40; of which 1/5 = 8; this is the length. Proof: Multiply the 2 1/2 schoenia of the width x [the 8] of the length [and it will be] 20.;[- - -];;(Frg. III, col. vii);The [monthly] interest [on 100 drachmai is 2 drachmai 3 obols]; let 4 months pass, and the whole sum [be ] 1100 drachmai; what is the [principal? Multiply] the 2 drachmai 3 obols into the [4 months,] = 10; into 1100 = 11000, [of which 1/100 =] 110 drachmai; I took 1/110 of 11,000 = 100; this is the interest on 1000 (drachmai); so much is the principal.;The [monthly] interest [on 100 drachmai] is 2 drachmai 3 obols; let 4 months pass, and the principal be 200 drachmai; what is the interest? [Multiply] the 2 drachmai 3 obols into the [4] months = 10; into 200 = 2000, of which 1/100 = 20; so much is the interest.;There are artabai of wheat [measured] by the 36-choinix measure; how many will there be by the 40-choinix measure? [Multiply] the 30 artabai of wheat by the 36 [=] 1080, of which 1/40 = 27. [There are so many by the 40-]choinix measure.;;(Frg. IV: too fragmentary to be translated);;(Frg. V);[- - -] the mile contains [8(?)] stadia [- - - the stadium] contains 4 [1/6] geometrical schoenia [- - -] and the mile contains [- - -];;(Frg. VI-X too fragmentary to be translated)