Aqoon aasaasi ah oo ku saabsan jajabyada: Fududaynta arrimaha la midka ah waa la oggol yahay. Fududaynta lambarrada la midka ah ma aha. Haddana, waxaa jira jajabyo halkaas oo fududeyntan mamnuuca ah ay u muuqato inay shaqeyso. Waxaa mudan in si dhow loo baaro qoyska jajabyada si gaar ah u fudud: jajabyada halkaas oo tiro isku mid ah ay ka muuqato dhammaadka tirinta iyo bilowga tirinta.
Lambarka waa inuu hadda ka koobnaadaa tiro \(a\) iyo tiro ku lifaaqan \(x\) , hooseeyaha isla lambar \(x\) iyo tiro ku lifaaqan \(c\):
$$\frac{\overline{a\,x}}{\overline{x\,c}}$$
Tirada guud ee lambarrada ku jira tiro-koobaha iyo hooseeyaha ha ahaato \(n \ge 2\) .
Markaas \(a\) iyo \(c\) mid walba wuxuu leeyahay \(k=n-1\) boosas jajab tobanle ah. Intaa waxaa dheer, ha u daa \(x \in \{1,\dots,9\}\) .
Qorista jajab tobanlaha caadiga ah, tani macnaheedu waa:
$$\overline{a\,x}\,=10a+x$$
iyo
$$\overline{x\,c}\,=x10^k+c.$$
Sidaa darteed, tirtirka mamnuuca ah, kaas oo halkan lagu baarayo, wuxuu noqon doonaa:
$$\frac{\overline{a\,x}}{\overline{x\,c}} \longmapsto \frac{a}{c}$$
Waxaan si sax ah u raadineynaa kiisaska ay qiimuhu uusan isbeddelin.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Qaab-dhismeedka oo dhan hadda waxaa laga soo saari karaa isbeddello fudud. Wax-soo-saarka isku-dhufashada...:
$$c(10a+x)=a(x10^k+c)$$
Marka la sii badiyo,:
$$10ac+cx=ax10^k+ac$$
Haddii aad si habboon u isku darto ereyada, waxaa soo socota in:
$$9ac=x(a10^k-c)$$
Maadaama \(a\) iyo \(c\) ay yihiin tiro kasta oo \(k\) ah oo togan, \(a10^kc>0\) . Tani waa sababta oo ah \(a10^k\) had iyo jeer way ka weyn tahay lambar kasta oo \(k\) ah oo \(c\) ah. Sidaa darteed, waan kala qaybin karnaa oo heli karnaa xaaladda dhexe.:
$$\boxed{x=\frac{9ac}{a10^k-c}}$$
Qaaciddadani waxay si sax ah u qeexaysaa marka fududaynta muuqata ay ka shaqeyso qaabkan gaarka ah. Ma aha oo kaliya lagama maarmaan laakiin sidoo kale waa ku filan tahay: Haddii isle'egtani ay jirto, markaa dhammaan isbeddellada waa la rogi karaa oo dib loogu celin karaa natiijadii asalka ahayd.:
$$\frac{10a+x}{x10^k+c}=\frac{a}{c}$$
Qodobka muhiimka ah ayaa ah in tibaaxda \(\frac{9ac}{a10^kc}\) ay ugu dambeyntii keento hal tiro jajab tobanle ah oo ka timid \(\{1,\dots,9\}\) . Markaas oo keliya ayaa jajab noocaas ah soo baxaa. Jajabyada saxda ah \(a<c\) ayaa sidoo kale loo baahan yahay. Markaas \(\frac{a}{c}<1\) sidoo kale waa run, sababtoo ah sinnaanta qiimayaasha, jajabka asalka ah sidoo kale waa sax.
Qaaciddada \(\displaystyle x=\frac{9ac}{a10^kc}\) aad bay ugu habboon tahay caddaynta. Si kastaba ha ahaatee, qaab dib loo habeeyay ayaa aad waxtar ugu leh helitaanka tusaalooyin noocaas ah. Laga soo bilaabo isla'egta \(\frac{10a+x}{x10^k+c}=\frac{a}{c}\) waxaan horey u helnay \(\displaystyle 9ac=x(a10^kc)\) . Si la mid ah, waxaan haynaa \(\displaystyle c(9a+x)=xa10^k\) .
Hadda waxaan isku daraa qaybiyaha guud ee \(a\) iyo \(x\) . Aan \(\displaystyle g=\gcd(a,x)\) . Kadib waxaa jira tirooyin \(b\) iyo \(y\) oo ah in \(\displaystyle a=gb\) , \(\displaystyle x=gy\) , iyo \(\displaystyle \gcd(b,y)=1\) . Adigoo tan ku beddelaya \(\displaystyle c(9a+x)=xa10^k\) waxaan helnaa \(\displaystyle c(9b+y)=x b10^k\) . Maadaama \(\displaystyle \gcd(9b+y,b)=\gcd(y,b)=1\) , qodobka \(9b+y\) waa inuu si buuxda u qaybiyaa tibaaxda \(\displaystyle x10^k\) . Haddii aan dejino \(\displaystyle d=9b+y\) , ka dibna \(\displaystyle d\mid x10^k\) iyo sidoo kale \(\displaystyle d\equiv y \pmod 9\) . Taas beddelkeeda, qaybiyeyaasha noocaas ah waxaan si toos ah uga heli karnaa
$$\displaystyle a=g\frac{d-y}{9}$$
iyo
$$\displaystyle c=\frac{x10^k(d-y)}{9d}.$$
Tani waxay meesha ka saaraysaa baahida loo qabo in si indho la'aan ah loo tijaabiyo \(a\) iyo \(c\) . Tiro kasta oo \(x\in\{1,\dots,9\}\) , qaybiye kasta oo \(g\mid x\) , iyo qaybiye kasta oo ku habboon \(d\mid x10^k\) musharrixiinta ayaa la helayaa. Waxa kaliya ee haray waa in la hubiyo in \(a\) iyo \(c\) ay runtii yihiin tirooyin \(k\) ah, iyo, haddii jajabyada saxda ah la rabo, inay yihiin \(a<c\) . Tani si weyn ayay u dhakhso badan tahay sababtoo ah \(\displaystyle 10^k=2^k5^k\) waxay leedahay qaybiyeyaal aad u yar marka loo eego tirada guud ee lambarrada \(k\) .
Laba tusaale:
$$\frac{16}{64}=\frac{1}{4}$$
Halkan, lambarka \(6\) waa la tirtiray.
Tusaale aad u dheer oo leh \(42\) oo jajab tobanle ah midkiiba iyo joojinta soo noqnoqda waa:
$$\frac{166666666666666666666666666666666666666666}{666666666666666666666666666666666666666664}=\frac{1}{4}$$
Halkan sidoo kale, isla lambarkii ayaa laga saarayaa: lambarkii ugu dambeeyay \(6\) , tirada hoose ee ugu horreysa \(6\) .