Kuthi Ngubani ofuna ukuba nguMillionaire kutshanje bekukho umbuzo omncinci omnandi ekuye kwafuneka uGünther Jauch acinge ngawo ngokucacileyo: "Inani lihlala lihlulwa ngo \(4\) ngaphandle kwentsalela ukuba inani lenziwe ukusuka kumanani alo amabini okugqibela...?" -kwaye kulapho kanye kufuneka ucinge ngemathematika okomzuzwana endaweni yokuzivumela ukuba utsalwe ziziphazamiso ezintle. Kuba ngelixa iimpendulo ezinje ngo “lingene”, “ziqulathe \(0\) ” okanye “isixa-mali esingumnqamlezo silingana \(4\) ” nesandi esikholelekayo xa usijonga nje kuqala, impendulo echanekileyo isekelwe kwipropathi elula yenkqubo yethu yokugqibela.
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.