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# Title: Leibniz-Handschriften zur Mathematik LH 35, 3 B 11
# Folio 9v, page id 00068261:0018
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in progreff Exprissione Denaria 2 - 1 p dibic habet ultimam notam 3, ergo
si per 31 exacte diviti potest, habebit quotiens ultimam notam 3
In Expressione Novenaria 2 -1 habet ultimam notam 0, et 31 novenarie est
34. Ergo 2 -1 divis. per 31. dabit quotientem cujus ultima nota novenarie est o
Ergo idem quoliens devarid expressus habet ultimam notam 3, et novenarie expressus
habet altimam notam. 0. seu idem numerus  quotiers q est  ex + 3
et
Denarie sit 2 = t = 3, et  2 - 1, : 9t -. M3 Fiel
D
Resid 3y = h
Rurgus a Resid  .. h3 =  Ergo h = 6
Rursus Rosid. nA62 = 7  Ergo . H63- 7 seu
K5B dinidi potest per 8. Cergont i. k dividi potest per 8. Barsus t63 = 0
Ergo fit 220 - 1=.. Y63
sed 3s duidi potest peqe B, s
sinos esis
seu. K 63 exacte dividi potest per 7 sed 63 exacte dividi potest, per 7. Ergo  rk poterit
urgo et K poterit pacte dividi per 7, sed ilam spacte dividi potest per 8. Ergo _ K = 6 nam monersus
numerus qui tam per 7 quam ser B dividi exacte potest necessario terminatur per 6
Burjus nR63  divisus per 6 relinquet 3. Ergo. K60 exacte dividi potest, u 6.
Ergo et iK exacte dividi potest per 6. Hinc quia.. K exacte dividi potest per 6,7, 8
 Sequitur quod exacte dividi possit per productum ex 3,7, 8, seu per 168
 in expressione octonaria 2  1 habet ultimam notam 7. Et
1+  31 ex primitur per 37. Ergo si dividas 2 - 1 per 31, qua
quotiens octonarie expressus habet ultimam notam 1
in expressione septenaria 2-1 habet ultimam notam 0, et
et 31 exprimitur per 43. Eq quod si 23 - dividatur per 31, quotiens
septenarie exprossus habet ultimam notam o
in expressione seraria 20-1 habet ultimam notam 3, eti
31 exprimitur per SI. Ergo quo si 370- 1 dividas per 31, quotiens
senane expressas habebit altimam notam, 3
Ia expressione quinaria 230 - 1 habet ultima ia  ) 99
si 31 dividas per 4, umerum periddi, restat 7, sed 3tia nota periodi est 4). Ergo si 230-
dividas per 31 quotiens quinanie expressus est 4
itaque  si 2 = 1 dividas per 31, quotiens sit 4, et 9 per 10 divisus
do
relinquet 3, per 9 divisus relinquel 0, per 8 divisus reliciquet 7, per 7 divisus
relinquet 0, per 6 divisus relinquet 3, per 5 divisus relinquet 4
- 30 -1 = bxt3~ cg 9c Y Xb + 3 = 9c = Bd +7 = 7ex = 6f + 3-59+ 4
Ergo  divisibile per 3. 1t d per 7 et e per 9 et f pM Z e per 3 (1 tandem = 31K