B\u00d6L\u00dcNEB\u0130LME<\/p>\n
2 \u0130le B\u00f6l\u00fcnebilme
\nx = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 2 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx \u00ba 0 (mod2) olmal\u0131
\nx = an.10n+an-1.10n-1+an-2.10n-2+ . . . +a1.101+a0
\n10 \u00ba 0(mod2) oldu\u011funa g\u00f6re “n\u2208Ni\u00e7in 10n\u00ba 0 (mod2)
\nx \u00ba 0+0+0+ . . . +a0 \u00ba 0 (mod2) olmal\u0131.<\/p>\n
Demek ki a0 \u00ba 0(mod2) olmal\u0131.<\/p>\n
O halde son basamaktaki say\u0131 \u00e7ift olmal\u0131d\u0131r.<\/p>\n
3 \u0130le B\u00f6l\u00fcnebilme<\/p>\n
x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 3 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx \u00ba 0 (mod3) olmal\u0131
\nx = an.10n+an-1.10n-1+an-2.10n-2+ . . . +a1.101+a0
\n10 \u00ba1 (mod3) oldu\u011funa g\u00f6re “n\u2208Ni\u00e7in 10n \u00ba 1(mod3)
\nx \u00ba an.1+an-1.1+ . . . +a.1+a0 \u00ba 0 (mod3) olmal\u0131<\/p>\n
Demek ki an+an-1+an-2+ . . . +a1+a0 \u00ba 0 (mod3) olmal\u0131 <\/p>\n
O halde rakamlar\u0131n\u0131n toplam\u0131 3 \u00fcn kat\u0131 olmal\u0131d\u0131r.<\/p>\n
4 \u0130le B\u00f6l\u00fcnebilme<\/p>\n
x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 4 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx = an.10n+an-1.10n-1+an-2.10n-2+ . . .+a2.102+a1.101+a0 \u00ba0 (mod4) olmal\u0131<\/p>\n
101 \u00ba 2 (mod4)
\n102 \u00ba 0 (mod4)
\n103 \u00ba 0 (mod4)
\n104 \u00ba 0 (mod4)<\/p>\n
O halde
\nx \u00ba an.0+an-1.0+ . . . +a2.0+a1.10+a0 \u00ba 0 (mod4)
\na1.10+a0 \u00ba 0 (mod4) olmal\u0131 <\/p>\n
O halde say\u0131n\u0131n son iki basama\u011f\u0131ndaki say\u0131 4 ile tam b\u00f6l\u00fcnebilmelidir. <\/p>\n
5 \u0130le B\u00f6l\u00fcnebilme <\/p>\n
x = anan-1an-2 . . .a0 say\u0131s\u0131n\u0131n 5 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx \u00ba 0 (mod5) olmal\u0131
\nx = an.10n+an-1.10n-1+an-2.10n-2+ . . .+a1.101+a0
\n10 \u00ba 0 (mod5) oldu\u011funa g\u00f6re “n\u2208Ni\u00e7in 10n\u00ba 0(mod5)
\nx \u00ba an.0+an-1.0+ . . . +a1.0+a0 \u00ba 0 (mod5) olmal\u0131
\na0 \u00ba (mod5)<\/p>\n
O halde son basamaktaki say\u0131 0 ya da 5 olmal\u0131d\u0131r.<\/p>\n
6 \u0130le B\u00f6l\u00fcnebilme
\nx = anan-1an-2 . . .a1a0 say\u0131s\u0131n\u0131n 6 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx = an.10n+an-1.10n-1+ . . . +a3.103+a2.102+a1.101+a0 \u00ba 0(mod6) olmal\u0131
\n6 = 2 . 3 oldu\u011funa g\u00f6re x \u00ba 0 (mod6) ise
\n x \u00ba 0 (mod2) ve x \u00ba 0 (mod3) olmal\u0131d\u0131r.<\/p>\n
O halde hem 2 ile hem de 3 ile b\u00f6l\u00fcnebilme kural\u0131n\u0131 birlikte sa\u011flamal\u0131d\u0131r.<\/p>\n
7 \u0130le B\u00f6l\u00fcnebilme <\/p>\n
x = anan-1an-2 . . .a1a0 say\u0131s\u0131n\u0131n 7 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx = an.10n+an-1.10n-1+ . . . +a3.103+a2.102+a1.101+a0 \u00ba 0(mod7)<\/p>\n
101 \u00ba 3 (mod7)
\n102 \u00ba 2 (mod7)
\n103 \u00ba 6 \u00ba -1 (mod7)
\n104 \u00ba-3 (mod7)
\n105 \u00ba-2 (mod7)
\n106 \u00ba 1 (mod7)<\/p>\n
x = . . . +a6.(1) + a5.(-2)+a4.(-3) + a3.(-1) + a2.2+a1.3+a0 = 0 (mod7) <\/p>\n
+ – +<\/p>\n
O halde say\u0131n\u0131n basamaklar\u0131n\u0131n sa\u011fdan sola do\u011fru 3\u2019er 3\u2019er gruplad\u0131ktan sonra her grup s\u0131ras\u0131yla birer birer (+) yada (-) i\u015faretleri koyulduktan sonra sa\u011fdan sola do\u011fru her basamaktaki say\u0131y\u0131 s\u0131ras\u0131yla i\u015faretleri ve \u201c1\u201d,\u201d3\u201d ve \u201c2\u201d say\u0131lar\u0131yla \u00e7arpt\u0131ktan sonra bulunan toplam say\u0131 7\u2019nin kat\u0131 olmal\u0131d\u0131r.<\/p>\n
8 \u0130le B\u00f6l\u00fcnebilme <\/p>\n
x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 8 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx \u00ba 0(mod8) olmal\u0131
\nx = an.10n+an-1.10n-1+ . . . +a3.103+a2.102+a1.101+a0 \u00ba 0(mod8) olmal\u0131<\/p>\n
101 \u00ba 2 (mod8)
\n102 \u00ba 4 (mod8)
\n103 \u00ba 0 (mod8) “n\u2208N+ ve n \u00b3 3 i\u00e7in 10n \u00ba 0 (mod8)
\n104 \u00ba 0 (mod8)<\/p>\n
x = an.0+an-1.0+ . . . + a3.0+a2.102+a1.10+a0 \u00ba 0 (mod8) olmal\u0131
\na2.102+a1.10+a0 = a2a1a0 \u00ba 0 (mod8) olmal\u0131<\/p>\n
O halde son 3 basama\u011f\u0131ndaki say\u0131 8 in kat\u0131 olmal\u0131d\u0131r.<\/p>\n
9 \u0130le B\u00f6l\u00fcnebilme <\/p>\n
x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 9 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in
\nx = an.10n+an-1.10n-1+ . . . +a3.103+a2.102+a1.101+a0 \u00ba 0 (mod9) olmal\u0131.
\n10 \u00ba 1(mod9) “n\u2208Ni\u00e7in 10n\u00ba 1(mod9)<\/p>\n
x = an.1+an-1.1+an-2.1+ . . . +a1.1+a0 \u00ba 0 (mod9) olur
\nan+an-1+an-2+ . . . a1+a0 \u00ba 0 (mod9) olur.<\/p>\n
O halde say\u0131n\u0131n rakamlar\u0131n\u0131n toplam\u0131 9\u2019un kat\u0131 olmal\u0131d\u0131r.<\/p>\n
11 \u0130le B\u00f6l\u00fcnebilme <\/p>\n
x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 11 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in <\/p>\n
x \u00ba 0 (mod11) olmal\u0131
\n x = an.10n+an-1.10n-1+ . . . +a3.103+a2.102+a1.101+a0<\/p>\n
“n\u2208Nve n, \u00e7ift i\u00e7in 10n \u00ba 1
\n“n\u2208Nve n, tek i\u00e7in 10n \u00ba-1<\/p>\n
101 \u00ba -1 (mod11)
\n102 =100 \u00ba 1 (mod11)
\n103 \u00ba-1 (mod11)
\n104 \u00ba 1 (mod11)
\n105 \u00ba-1 (mod11)
\n106 \u00ba 1 (mod11)<\/p>\n
x = an.(1)+an-1.(-1)+an-2.(1)+ . . . +a2.(1)+a1.(-1)+a0
\nan-an-1+an-2+ . . . +a2-a1+a0 \u00ba 0 (mod11)<\/p>\n
O halde say\u0131n\u0131n rakamlar\u0131 sa\u011fdan sola do\u011fru (+1) ve (-1) ile \u00e7arparak topland\u0131\u011f\u0131nda bulunan say\u0131 11\u2019in kat\u0131 olmal\u0131d\u0131r.<\/p>\n
21 \u0130le B\u00f6l\u00fcnebilme<\/p>\n
21 = 3 . 7
\nHem 3 hem de 7 ile b\u00f6l\u00fcnebilme kurallar\u0131n\u0131 birlikte sa\u011flamal\u0131d\u0131r.<\/p>\n","protected":false},"excerpt":{"rendered":"
B\u00d6L\u00dcNEB\u0130LME 2 \u0130le B\u00f6l\u00fcnebilme x = anan-1an-2 . . . a0 say\u0131s\u0131n\u0131n 2 ile tam b\u00f6l\u00fcnebilmesi i\u00e7in x \u00ba 0 (mod2) olmal\u0131 x = an.10n+an-1.10n-1+an-2.10n-2+ . . . +a1.101+a0 10 \u00ba 0(mod2) oldu\u011funa g\u00f6re “n\u2208Ni\u00e7in 10n\u00ba 0 (mod2) x \u00ba 0+0+0+ . . . +a0 \u00ba 0 (mod2) olmal\u0131. Demek ki a0 \u00ba 0(mod2) olmal\u0131. …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7343,7335,7344,7336,7337,7338,7339,7340,7341,7342,7334],"class_list":["post-3126","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-11-ile-bolunebilme","tag-2-ile-bolunebilme","tag-21-ile-bolunebilme","tag-3-ile-bolunebilme","tag-4-ile-bolunebilme","tag-5-ile-bolunebilme","tag-6-ile-bolunebilme","tag-7-ile-bolunebilme","tag-8-ile-bolunebilme","tag-9-ile-bolunebilme","tag-bolunebilme-kurallari"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3126","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=3126"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3126\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3126"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3126"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3126"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}