Kwisiqendu sakutshanje se- "Who Wants to Be a Millionaire ?", bekukho umbuzo omncinci ocacileyo uGünther Jauch ekufuneka acinge ngawo: "Inani lihlala lahlulwa ngo \(4\) ngaphandle kwentsalela ukuba inani elenziwe kwiinombolo zalo ezimbini zokugqibela...?" - kwaye kulapho kanye kufuneka ucinge ngokwezibalo okomzuzwana, endaweni yokutsalwa ziziphazamiso ezintle. Kuba nangona iimpendulo ezinje ngo-"zilingana", "ziqulethe u \(0\) , okanye "isimbuku seenombolo ngu \(4\) " zivakala zinengqondo xa uzijonga okokuqala, impendulo echanekileyo ikwipropathi elula yenkqubo yethu yedesimali.
Inani \(X\) liyahlulwahlulwa ngo \(4\) ukuba kwaye kuphela ukuba inani elenziwe ngamanani alo amabini okugqibela liyahlulwa ngo \(4\) . Ubungqina bulandela ngqo kwisimo sedesimali. Inani ngalinye lendalo \(X\) lingamelwa ngokukodwa kwimo
\[
X = 100 \cdot X' + X''
\]
bhala apho \(X''\) linani elenziwe ngamanani amabini okugqibela, oko kukuthi, \(0 \leq X'' < 100\) , kunye \(X'\) yinxalenye eyandulelayo yenani.
\[
100 = 25 \cdot 4
\]
iyasebenza, ilandela
\[
X = 25 \cdot 4 \cdot X' + X''.
\]
Isihlomelo sokuqala \(25 \cdot 4 \cdot X'\) sihlala sahlulwahlulwa ngo \(4\) nokuba \(X'\) . Ke ngoko, kwintsalela ye \(X\) xa yahlulwe ngo \(4\) kuphela \(X''\) efanelekileyo. Ichazwe ngokusesikweni:
\[
X \equiv X'' \pmod{4}.
\]
Oku kusebenza ngokukodwa:
\[
4 \mid X \iff 4 \mid X''.
\]
Imithetho efanayo yokwahlulwahlulwa ivela nanini na xa amandla e- \(10\) modulo inani liba lula kakhulu. Ukuze ukwahlulwahlulwahlulwa ngo \(4\) into ebalulekileyo yayikukuba i \(100 \equiv 0 \pmod 4\) Kuba nomdla ngakumbi xa amaxabiso \(1\) okanye \(-1\) esenzeka endaweni ye \(0\) .
Umzekelo oqhelekileyo kukuhlukana nge \(11\) .
\[
10 \equiv -1 \pmod{11}
\]
Ukuba oku kuyinyani, amaxabiso endawo e-modulo yenani ledesimali elingu \(11\) ahlala etshintshana phakathi kuka \(1\) no \(-1\) .
\[
X = a_0 + 10a_1 + 10^2a_2 + 10^3a_3 + \dots
\]
Ngoko ke kulandela
\[
X \equiv a_0 - a_1 + a_2 - a_3 + \dots \pmod{11}.
\]
Inani liyahlulwa ngo- \(11\) ukuba kuphela xa isimbuku samanani alo atshintshanayo sihlulwa ngo- \(11\) Umzekelo, ku- \(918082\) oku kunjalo.
\[
2 - 8 + 0 - 8 + 1 - 9 = -22,
\]
kwaye ekubeni \(-22\) yahlulwe yi \(11\) , \(918082\) yahlulwe yi \(11\) .
Eyona nto intle ngakumbi ngumthetho we- \(7\) , \(11\) kunye ne- \(13\) ngaxeshanye. Uqinisekisa ukuba
\[
1001 = 7 \cdot 11 \cdot 13
\]
kwaye njalo
\[
1000 \equiv -1 \pmod{7}, \qquad
1000 \equiv -1 \pmod{11}, \qquad
1000 \equiv -1 \pmod{13}.
\]
Ukuba wahlula inani libe ziibhloko ezintathu ukusuka ekunene ukuya ekhohlo, ngoko ke ungadibanisa kwaye uthabathe ezi bhloko ngokwahlukana.
\[
X = 123456789
\]
Ngoko ke, ukuba umntu ucinga
\[
789 - 456 + 123 = 456.
\]
Inani lokuqala linemodulo efanayo eseleyo engu \(7\) , \(11\) kunye no \(13\) njenge- \(456\) Ke ngoko, inani elikhulu kakhulu linokutshintshwa lelinye elincinci kakhulu ngaphandle kokutshintsha ukwahlulwa kwalo ngala manani mathathu.
Ukuhlukana nge \(37\) nako kunesimo esihle ngendlela emangalisayo.
\[
999 = 27 \cdot 37
\]
iyasebenza
\[
1000 \equiv 1 \pmod{37}.
\]
Xa zahlulwe ngama- \(37\) iibhloko ezintathu zingongezwa kunye. Umzekelo, ukusuka
\[
99937
\]
isixa-mali
\[
99 + 937 = 1036.
\]
Pha
\[
1036 = 28 \cdot 37
\]
Ukuba \(99937\) iyahlulwahlulwa ngo-37, ngoko ke i-99937 nayo iyahlulwahlulwa ngo- \(37\) .
Ekuqaleni imithetho enjalo ibonakala ngathi ngamaqhinga eenombolo, kodwa ekugqibeleni zizicelo nje zengcamango efanayo: ukutshintsha amandla amakhulu e-10 nge-simple residuals modulo inani elithethwayo. Oku kuguqula inani elikhulu ledesimali libe yi-calculation elula equka ii-congruences. Yiyo loo nto imithetho yokwahlulwahlulwa ingaphezulu kwamaqhinga nje e-arithmetic; imele ukuncitshiswa kwi-residues modulo. \(10^k\): Kwifomathi yemibuzo, zibonakala njengemigibe emincinci yokuqonda, kodwa zikhokelela ngqo kwingcamango emangalisa ngokumangalisayo kwithiyori yamanani.