{"id":49603,"date":"2014-02-08T14:10:19","date_gmt":"2014-02-08T12:10:19","guid":{"rendered":"http:\/\/www.sorubak.com\/blog\/?p=49603"},"modified":"2021-11-27T13:37:30","modified_gmt":"2021-11-27T10:37:30","slug":"okek-ve-cozumlu-sorular","status":"publish","type":"post","link":"https:\/\/www.sorubak.com\/blog\/okek-ve-cozumlu-sorular.html","title":{"rendered":"OKEK ve \u00c7\u00f6z\u00fcml\u00fc Sorular"},"content":{"rendered":"

OKEK (ORTAK KATLARIN EN K\u00dc\u00c7\u00dc\u011e\u00dc)<\/strong><\/p>\n

\u0130ki veya daha \u00e7ok say\u0131n\u0131n her birine b\u00f6l\u00fcnen en k\u00fc\u00e7\u00fck say\u0131d\u0131r. Verilen iki veya daha \u00e7ok say\u0131n\u0131n OKEK\u2019 ini bulmak i\u00e7in, say\u0131lar asal \u00e7arpanlar\u0131n\u0131n kuvvetleri cinsinden yaz\u0131l\u0131r ve ortak asal \u00e7arpanlar\u0131ndan \u00fcsleri en b\u00fcy\u00fck olanlarla ortak olmayan asal \u00e7arpanlar\u0131n\u0131n t\u00fcm\u00fc al\u0131narak \u00e7arp\u0131l\u0131r.<\/p>\n

1. Aralar\u0131nda asal say\u0131lar\u0131n OKEK\u2019 i, bu say\u0131lar\u0131n \u00e7arp\u0131mlar\u0131na e\u015fittir. Yani, a ile b say\u0131s\u0131 aralar\u0131nda asal say\u0131lar ise,
\n(a, b)OKEK = a . b dir.
\n2. a ve b iki do\u011fal say\u0131 olmak \u00fczere, bu iki do\u011fal say\u0131n\u0131n OBEB\u2019 i ile OKEK\u2019 inin \u00e7arp\u0131m\u0131, bu iki do\u011fal say\u0131n\u0131n \u00e7arp\u0131m\u0131na e\u015fittir. Yani, a ve b do\u011fal say\u0131s\u0131 i\u00e7in
\na . b = (a, b)OKEK . (a, b)OBEB dir.
\n3. a, b, c, d sayma say\u0131lar\u0131 olmak \u00fczere,
\n(a\/c,b\/d)OKEK = (a, b)OKEK \/ (c, d)OBEB dir.
\n4. a ve b iki do\u011fal say\u0131 olmak \u00fczere,
\n(a, b)OKEK = x ve (a, b)OBEB = y
\nise, a ile b say\u0131lar\u0131n\u0131n toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011feri
\nx + y dir.
\n5. Ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131n\u0131n OKEK\u2019 i bu iki say\u0131n\u0131n \u00e7arp\u0131m\u0131na e\u015fittir. Yani, a ile b ard\u0131\u015f\u0131k iki sayma say\u0131s\u0131 olmak \u00fczere,
\n(a, b)OKEK = a . b dir.
\n6. a ile b sayma say\u0131lar\u0131 olmak \u00fczere, a < b ise,
\n(a, b)OBEB <= a <= b <= (a, b)OKEK dir.<\/p>\n

\u00d6rnek :<\/strong><\/p>\n

18 ile 45 say\u0131lar\u0131n\u0131n OKEK\u2019 ini bulunuz.<\/p>\n

\u00c7\u00f6z\u00fcm:<\/strong><\/p>\n

18 = 2 . 32
\n45 = 32 . 5
\noldu\u011fundan, (18, 45)OKEK = 32 . 2 . 5 = 90 olur.<\/p>\n

\u00d6rnek :<\/strong><\/p>\n

a ve b do\u011fal say\u0131lar\u0131n\u0131n OKEK\u2019 i 48 ve OBEB\u2019 i 8 ve bu say\u0131lardan biri 16 ise, di\u011fer say\u0131 ka\u00e7t\u0131r?<\/p>\n

\u00c7\u00f6z\u00fcm:<\/strong>
\na = 16 olsun. (16, b)OKEK = 48 ve (16, b)OBEB = 8 oldu\u011funa g\u00f6re,
\na . b = (a, b)OKEK . (a, b)OBEB
\n16 . b = 48 . 8
\nb = 24
\nbulunur.<\/p>\n

\u00d6rnek :<\/strong><\/p>\n

Herhangi iki do\u011fal say\u0131n\u0131n OKEK\u2019 i 120 ve OBEB\u2019 i 8 oldu\u011funa g\u00f6re, bu say\u0131lar\u0131n toplam\u0131 en \u00e7ok ka\u00e7 olabilir?<\/p>\n

\u00c7\u00f6z\u00fcm:<\/strong>
\n\u0130ki do\u011fal say\u0131n\u0131n toplam\u0131 en \u00e7ok bu iki say\u0131n\u0131n OBEB\u2019 ile OKEK\u2019 inin toplam\u0131 kadar olabilece\u011finden,<\/p>\n

120 + 8 = 128 dir.
\n\u00d6rnek :<\/p>\n

Boyutlar\u0131 2 cm, 4 cm, 6 cm olan dikd\u00f6rtgenler prizmas\u0131 bi\u00e7imindeki kutunun i\u00e7erisi, bo\u015f yer kalmayacak \u015fekilde en k\u00fc\u00e7\u00fck boyutlu k\u00fcplerle doldurulmak istenmektedir. Bu kutuya ka\u00e7 tane k\u00fcp yerle\u015ftirilebilir?<\/p>\n

\u00c7\u00f6z\u00fcm:<\/p>\n

Kutu en k\u00fc\u00e7\u00fck boyutlu k\u00fcplerle doldurulmak istendi\u011finden, 2 cm, 4 cm, 6 cm say\u0131lar\u0131n\u0131n OKEK\u2019 i bulunmal\u0131d\u0131r. Bu nedenle,
\n(2, 4, 6)OKEK = 12 t\u00fcr. B\u00f6ylece, en k\u00fc\u00e7\u00fck boyutlu k\u00fcp\u00fcn bir kenar\u0131 = 12 cm olur. Bir kenar\u0131 12 cm olacak \u015fekilde yerle\u015ftirilebilecek k\u00fcp say\u0131s\u0131,
\nK\u00fcp say\u0131s\u0131 = Kutunun hacmi \/ K\u00fcp\u00fcn hacmi = 12.12.12\/2.4.6 = 6.3.2 = 36
\ntane olur.<\/p>\n

\u00d6rnek :<\/p>\n

a, b, c asal say\u0131lar olmak \u00fczere,
\nx = a2 . b3 . c5 ve y = a5 . c2
\nise, (x, y)OBEB = ? ve (x, y)OKEK = ? bulunuz.<\/p>\n

\u00c7\u00f6z\u00fcm:
\n(x, y)OBEB = a2 . c2 = (a . c)2
\n(x, y)OKEK = a5 . b3 . c5 olur.<\/p>\n

\u00d6rnek :<\/p>\n

Ay\u015fe toplar\u0131n\u0131 2\u2032 \u015fer 2\u2032 \u015fer, 4\u2032 er 4\u2032 er, 6\u2032 \u015far 6\u2032 \u015far sayarsa, her defas\u0131nda 1 top art\u0131yor. Ay\u015fe\u2019 nin en az ka\u00e7 topu vard\u0131r?<\/p>\n

\u00c7\u00f6z\u00fcm:
\nTop say\u0131s\u0131 = (2, 4, 6)OKEK + 1 = 12 + 1 = 13 t\u00fcr.<\/p>\n

\u00d6rnek :<\/p>\n

2, 3, 4 say\u0131lar\u0131na b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde 1 kalan\u0131n\u0131 veren en b\u00fcy\u00fck 2 basamakl\u0131 do\u011fal say\u0131 ka\u00e7t\u0131r?<\/p>\n

\u00c7\u00f6z\u00fcm:
\n[(2, 3, 4)OKEK] . k + 1 <= 99
\n24 . k + 1 <= 99
\nk = 4 olur. Buradan, say\u0131
\n24 . 4 + 1 = 96 + 1 = 97
\nbulunur.<\/p>\n

\u00d6rnek :
\n\u0130ki yang\u0131n sireni 5\/7, 7\/8 saat aral\u0131klarla alarm vermektedirler. Bu iki yang\u0131n sireni ayn\u0131 anda en son Cuma g\u00fcn\u00fc sabah 04.00\u2032 de alarm verdiklerine g\u00f6re, hangi g\u00fcn saat ka\u00e7ta tekrar birlikte alarm verirler?<\/p>\n

\u00c7\u00f6z\u00fcm:<\/p>\n

Yang\u0131n sirenleri 5\/7, 7\/8 say\u0131lar\u0131n\u0131n OKEK\u2019 lerinde ayn\u0131 anda alarm verirler. Dolay\u0131s\u0131yla,
\n(5\/7, 7\/8)OKEK = (5, 7)OKEK \/ (7, 8)OBEB = 35 \/ 1 = 35 saat
\nsonra tekrar alarm verirler. O halde, Cumartesi g\u00fcn\u00fc saat 15.00\u2032 de tekrar alarm vereceklerdir.<\/p>\n

\u00d6rnek :
\nBir a do\u011fal say\u0131s\u0131 5\/3, 6 say\u0131lar\u0131na b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde sonu\u00e7 tamsay\u0131 oldu\u011funa g\u00f6re, bu ko\u015fula uyan en k\u00fc\u00e7\u00fck a say\u0131s\u0131 ka\u00e7t\u0131r?<\/p>\n

\u00c7\u00f6z\u00fcm:
\n5\/3 ile 6\u2032 n\u0131n OKEK\u2019 ini bulmal\u0131y\u0131z. Bu takdirde,
\n(5\/3, 6)OKEK = (5, 6)OKEK \/ (3, 1)OBEB = 30 \/ 1 = 30 olur.<\/p>\n

\u00d6rnek :<\/p>\n

OKEK\u2019 i 7 olan a ve b do\u011fal say\u0131lar\u0131n\u0131n toplamlar\u0131n\u0131n en k\u00fc\u00e7\u00fck ve en b\u00fcy\u00fck de\u011ferlerinin \u00e7arp\u0131m\u0131 ka\u00e7 olur?<\/p>\n

\u00c7\u00f6z\u00fcm:<\/p>\n

(a, b)OKEK = 7 ve say\u0131lar\u0131n farkl\u0131 olmad\u0131klar\u0131 s\u00f6ylenmedi\u011fine g\u00f6re,
\na = 7 ve b = 7
\nal\u0131nabilir. Bu durumda, a ile b\u2019 nin toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011feri
\na + b = 7 + 7 = 14 \u2026 (1)
\nolur. Di\u011fer taraftan,
\na = 1 ve b = 7 al\u0131n\u0131rsa, a ile b\u2019 nin toplam\u0131n\u0131n en k\u00fc\u00e7\u00fck de\u011feri
\na + b = 1 +7 = 8 \u2026 (2)
\nolur. Buradan, (1) ile (2) nin \u00e7arp\u0131m\u0131 14 . 8 = 112 bulunur<\/p>\n","protected":false},"excerpt":{"rendered":"

OKEK (ORTAK KATLARIN EN K\u00dc\u00c7\u00dc\u011e\u00dc) \u0130ki veya daha \u00e7ok say\u0131n\u0131n her birine b\u00f6l\u00fcnen en k\u00fc\u00e7\u00fck say\u0131d\u0131r. Verilen iki veya daha \u00e7ok say\u0131n\u0131n OKEK\u2019 ini bulmak i\u00e7in, say\u0131lar asal \u00e7arpanlar\u0131n\u0131n kuvvetleri cinsinden yaz\u0131l\u0131r ve ortak asal \u00e7arpanlar\u0131ndan \u00fcsleri en b\u00fcy\u00fck olanlarla ortak olmayan asal \u00e7arpanlar\u0131n\u0131n t\u00fcm\u00fc al\u0131narak \u00e7arp\u0131l\u0131r. 1. Aralar\u0131nda asal say\u0131lar\u0131n OKEK\u2019 i, bu say\u0131lar\u0131n … Devam\u0131n\u0131 oku…<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13],"tags":[],"class_list":["post-49603","post","type-post","status-publish","format-standard","hentry","category-mt"],"_links":{"self":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/49603","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/comments?post=49603"}],"version-history":[{"count":0,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/posts\/49603\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/media?parent=49603"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/categories?post=49603"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.sorubak.com\/blog\/wp-json\/wp\/v2\/tags?post=49603"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}