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
$$\displaystyle c=\frac{x10^k(d-y)}{9d}.$$
Taas macnaheedu waa inaadan mar dambe indho la'aan noqon doonin. \(a\) iyo \(c\) Isku day. Lambar kasta. \(x\in\{1,\dots,9\}\), qaybiye kasta \(g\mid x\) iyo qaybiye kasta oo ku habboon \(d\mid x10^k\) Qofku wuxuu helaa musharrixiinta. Waxa kaliya ee haray waa in la hubiyo in \(a\) iyo \(c\) runtii \(k\)Waa tirooyin - iyo, haddii aad rabto jajabyo dhab ah, haddii \(a\) iyo \(c\) runtii \(k\)Waa tirooyin - iyo, haddii aad rabto jajabyo dhab ah, haddii \(a<c\) waxay khuseysaa.
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 \(6\) .