Esiqeshini sakamuva se- "Who Wants to Be a Millionaire ?", kwakunombuzo omncane ocacile uGünther Jauch okwakufanele awucabangele ngokusobala: "Inombolo ihlala ihlukaniswa ngo- \(4\) ngaphandle kwensalela uma inombolo eyakhiwe ngezinombolo zayo ezimbili zokugcina ingu...?" - futhi yilapho kanye lapho kufanele ucabange khona ngezibalo okwesikhashana, esikhundleni sokuhehwa yiziphazamiso ezinhle. Ngoba ngenkathi izimpendulo ezinjengokuthi "zilingana", "ziqukethe u \(0\) , noma "isamba sezinombolo singu \(4\) " zizwakala zinengqondo ekuqaleni, impendulo efanele itholakala endaweni elula yesistimu yethu yedesimali.
Inombolo \(X\) iyahlukaniswa ngo \(4\) uma futhi kuphela uma inombolo eyakhiwe ngamadijithi ayo amabili okugcina ihlukaniswa ngo \(4\) . Ubufakazi bulandela ngqo ekumelweni kwedesimali. Inombolo ngayinye yemvelo \(X\) ingamelwa ngendlela ehlukile efomini
\[
X = 100 \cdot X' + X''
\]
bhala ukuthi i \(X''\) iyinombolo eyakhiwe ngezinombolo ezimbili zokugcina, okungukuthi, \(0 \leq X'' < 100\) , kanye ne \(X'\) ingxenye eyandulele yenombolo. Kusukela
\[
100 = 25 \cdot 4
\]
kuyasebenza, kulandela
\[
X = 25 \cdot 4 \cdot X' + X''.
\]
Isithasiselo sokuqala \(25 \cdot 4 \cdot X'\) sihlala sihlukaniswa ngo \(4\) kungakhathaliseki ukuthi \(X'\) . Ngakho-ke, kokusele kuka \(X\) uma kuhlukaniswa ngo \(4\) kuphela \(X''\) okufanele. Kuvezwe ngokusemthethweni:
\[
X \equiv X'' \pmod{4}.
\]
Lokhu kusebenza ikakhulukazi:
\[
4 \mid X \iff 4 \mid X''.
\]
Imithetho efanayo yokuhlukaniswa ivela noma nini lapho amandla e-modulo \(10\) inombolo iba lula kakhulu. Ngokuhlukaniswa ngo \(4\) isici esibalulekile kwakuwukuthi \(100 \equiv 0 \pmod 4\) Kuba mnandi nakakhulu uma amanani \(1\) noma \(-1\) evele esikhundleni sika \(0\) .
Isibonelo esivamile ukuhlukaniswa ngo- \(11\) .
\[
10 \equiv -1 \pmod{11}
\]
Uma lokhu kuyiqiniso, amanani endawo e-modulo yenombolo yedesimali \(11\) ahlala eshintshana phakathi kuka \(1\) no \(-1\) .
\[
X = a_0 + 10a_1 + 10^2a_2 + 10^3a_3 + \dots
\]
Ngakho-ke kulandela
\[
X \equiv a_0 - a_1 + a_2 - a_3 + \dots \pmod{11}.
\]
Inombolo iyahlukaniswa ngo- \(11\) uma kuphela uma isamba sezinombolo zayo ezishintshanayo sihlukaniswa ngo- \(11\) Ku- \(918082\) isibonelo, lokhu kunjalo.
\[
2 - 8 + 0 - 8 + 1 - 9 = -22,
\]
futhi njengoba \(-22\) ihlukaniswa ngu- \(11\) , \(918082\) nayo ihlukaniswa ngu- \(11\) .
Okuhle nakakhulu umthetho we- \(7\) , \(11\) kanye ne- \(13\) ngasikhathi sinye. Uphethe lokho
\[
1001 = 7 \cdot 11 \cdot 13
\]
futhi kanjalo
\[
1000 \equiv -1 \pmod{7}, \qquad
1000 \equiv -1 \pmod{11}, \qquad
1000 \equiv -1 \pmod{13}.
\]
Uma uhlukanisa inombolo ibe amabhlogo amathathu ukusuka kwesokudla kuya kwesobunxele, ungangeza futhi ususe la mabhlogo ngokushintshana.
\[
X = 123456789
\]
Ngakho-ke, uma umuntu ecabanga
\[
789 - 456 + 123 = 456.
\]
Inombolo yokuqala ine-modulo efanayo esele engu \(7\) , \(11\) no \(13\) njengo- \(456\) Ngakho-ke, inombolo enkulu kakhulu ingathathelwa indawo yileyo encane kakhulu ngaphandle kokushintsha ukuhlukaniswa kwayo yilezi zinombolo ezintathu.
Ukuhlukaniswa ngo- \(37\) nakho kunesimo esihle ngokumangalisayo.
\[
999 = 27 \cdot 37
\]
kusebenza
\[
1000 \equiv 1 \pmod{37}.
\]
Uma ihlukaniswa ngo \(37\) amabhlogo amathathu angangezwa ndawonye. Isibonelo, kusukela ku-
\[
99937
\]
isamba
\[
99 + 937 = 1036.
\]
Lapho
\[
1036 = 28 \cdot 37
\]
Uma \(99937\) ehlukaniswa ngo-37, khona-ke u-99937 naye uhlukaniswa ngo- \(37\) .
Imithetho enjalo ekuqaleni ibonakala ingamaqhinga ezinombolo, kodwa ekugcineni imane iyizicelo zomqondo ofanayo: ukufaka esikhundleni samandla amakhulu eshumi nge-modulo elula ye-remnants inombolo okukhulunywa ngayo. Lokhu kuguqula inombolo enkulu yedesimali ibe ukubala okulula okubandakanya ukuhambisana. Yingakho imithetho enjalo yokuhlukanisa ingaphezu nje kwamaqhinga ezibalo; imelela ukwehliswa kwe-modulo ye-remnants \(10^k\): Ngefomethi yemibuzo, zibonakala njengezicupho ezincane zokuqonda, kodwa ziholela ngqo emcabangweni omuhle ngokumangalisayo kumbono wezinombolo.