{"id":6102,"date":"2013-10-21T15:01:30","date_gmt":"2013-10-21T21:01:30","guid":{"rendered":"http:\/\/19.org\/?p=6102"},"modified":"2013-11-10T23:54:11","modified_gmt":"2013-11-11T05:54:11","slug":"ikizkod-2","status":"publish","type":"post","link":"https:\/\/19.org\/tr\/ikizkod-2\/","title":{"rendered":"19 + Ba\u015fka bir say\u0131"},"content":{"rendered":"

19 say\u0131s\u0131ndan s\u00f6z eden tek sure olan 74’\u00fcnc\u00fc s\u0131rada yer alan M\u00fcddessir suresindeki gayb\u00ee haberin bir tecellisi olarak 1974 y\u0131l\u0131nda ke\u015ffedilen 19 sisteminin ke\u015ffinden sonra \u015feytan insanlar\u0131 bu mucizeye k\u00f6r ve sa\u011f\u0131r etmek i\u00e7in \u00e7e\u015fitli y\u00f6ntemler ve bahaneler kulland\u0131. \u00c7e\u015fitli yalanlar ve iftiralara ek olarak bir de say\u0131lar\u0131 suistimal etmeyi denedi. Rakamlarla keyfi ve tutars\u0131z hesaplar yaparak “bak ben de mucize ke\u015ffettim” diye ortal\u0131\u011fa f\u0131rlayan saf insanlar veya megalomanyaklar \u015feytan taraf\u0131ndan ba\u015far\u0131yla kullan\u0131ld\u0131lar ve kullan\u0131lmaya devam edilecekler… Liste uzun. T\u0131p doktoru Orhan Kuntman, Bahattin Uzunkaya, \u00d6mer \u00c7elak\u0131l, Uyduruk Hadis doktoru Halis Aydemir ve \u0130mran Akdemir bu \u00e7arp\u0131tmalar\u0131n\u0131 kitapla\u015ft\u0131ran T\u00fcrkiyeli isimler… \u00a0(Bu ki\u015filer dini konularda ayn\u0131 fikirleri payla\u015fm\u0131yorlar. \u00d6rne\u011fin \u0130mran karde\u015fimiz Hadis ve S\u00fcnneti Kuran’a ortak ko\u015fmuyor) Tar\u0131k Arabac\u0131 arkada\u015f\u0131m\u0131z a\u015fa\u011f\u0131daki makalede \u0130mran Akdemir’in \u0130kizkod diye ortal\u0131\u011fa s\u00fcrd\u00fc\u011f\u00fc kitap dolusu hesaplar\u0131n istatistiksel bir de\u011feri olmad\u0131\u011f\u0131n\u0131 sergiliyor. Bu konuda yazd\u0131\u011f\u0131m iki makale i\u00e7in bak:<\/p>\n

http:\/\/19.org\/tr\/2355\/ikizkod\/<\/a><\/p>\n

http:\/\/19.org\/tr\/2781\/simetrik\/<\/a><\/p>\n

\"\"<\/p>\n

\u0130mran Akdemir’in 19 ile 7’yi kar\u0131\u015ft\u0131rarak olu\u015fturdu\u011fu \u00f6rneklerden birisi.<\/p>\n

Sadece 19 vs 19 + Bir\u015fey<\/b><\/span><\/p>\n

Tar\u0131k Arabac\u0131
\n14 Ekim 2013<\/p>\n

B\u0130RKA\u00c7 AR\u0130TMET\u0130K SORUSU:<\/strong><\/p>\n

1) 1’den 1000’e kadar olan say\u0131lar\u0131n ka\u00e7 tanesi 19’un kat\u0131d\u0131r?<\/span>
\n2) 1’den 1000’e kadar olan say\u0131lar\u0131n ka\u00e7 tanesi 7’nin kat\u0131d\u0131r?<\/span>
\n3) 1’den 1000’e kadar olan say\u0131lar\u0131n ka\u00e7 tanesi biri 19’un di\u011feri 7’nin kat\u0131 olan iki do\u011fal say\u0131n\u0131n toplam\u0131 ya da fark\u0131 \u015feklinde yaz\u0131labilir?<\/span>
\n4) 1’den 1000’e kadar olan say\u0131lar\u0131n ka\u00e7 tanesi biri 19’un di\u011feri 7’nin kat\u0131 olan iki do\u011fal say\u0131n\u0131n toplam\u0131 \u015feklinde yaz\u0131labilir?<\/span><\/p>\n

OLASILI\u011eA D\u00d6KERSEK…<\/span><\/strong><\/p>\n

1) 1’den 1000’e kadar olan say\u0131lar\u0131n yakla\u015f\u0131k y\u00fczde ka\u00e7\u0131 19’un kat\u0131d\u0131r?<\/span>
\n2) 1’den 1000’e kadar olan say\u0131lar\u0131n yakla\u015f\u0131k y\u00fczde ka\u00e7\u0131 7’nin kat\u0131d\u0131r?<\/span>
\n3) 1’den 1000’e kadar olan say\u0131lar\u0131n yakla\u015f\u0131k y\u00fczde ka\u00e7\u0131 biri 19’un di\u011feri 7’nin kat\u0131 olan iki do\u011fal say\u0131n\u0131n toplam\u0131 ya da fark\u0131 \u015feklinde yaz\u0131labilir?<\/span>
\n4) 1’den 1000’e kadar olan say\u0131lar\u0131n yakla\u015f\u0131k y\u00fczde ka\u00e7\u0131 biri 19’un di\u011feri 7’nin kat\u0131 olan iki do\u011fal say\u0131n\u0131n toplam\u0131 \u015feklinde yaz\u0131labilir?<\/span><\/p>\n

Cevaplar s\u0131ras\u0131yla %5, %14, %81 ve %76 oluyor.<\/span><\/strong><\/span><\/p>\n

Sonu\u00e7 olarak 1,2 ve 3’teki olas\u0131l\u0131klar\u0131n toplam\u0131 %100’e tamamlan\u0131yor Yani -\u00e7\u0131karma yapmak serbestse- istisnas\u0131z HER SAYI “ya 19’un tamkat\u0131 ya 7’nin tamkat\u0131 ya da 19 ve 7’nin katlar\u0131n\u0131n toplam\u0131\/fark\u0131” \u015feklindedir.<\/span><\/p>\n

E\u011fer 4. sorudaki gibi \u00e7\u0131karma i\u015flemini yasaklarsak yine \u00e7ok bir\u015fey de\u011fi\u015fmiyor. Bu durumda (114’den k\u00fc\u00e7\u00fck) 54 tane say\u0131y\u0131 elde edemeyiz sadece. O kadar.<\/span><\/p>\n

1’den 1000’e kadar olan say\u0131lar\u0131n %95’i, (veya 114’den b\u00fcy\u00fck say\u0131lar\u0131n tamam\u0131) “ya 19’un tamkat\u0131 ya 7’nin tamkat\u0131 ya da 19 ve 7’nin pozitif katlar\u0131n\u0131n toplam\u0131” \u015feklindedir.<\/span><\/p>\n

Rastgele bir say\u0131n\u0131n tek ba\u015f\u0131na 19’un kat\u0131 olmas\u0131 d\u00fc\u015f\u00fck bir olas\u0131l\u0131kken, 19 ve 7’nin katlar\u0131 olmas\u0131 s\u0131radan bir matematiksel \u00f6zellik.<\/span><\/p>\n

Tabii ben mahsustan 19 ve 7 diye sordum. Aralar\u0131nda asal (ortak b\u00f6leni olmayan) herhangi iki asal say\u0131 i\u00e7in de durum bu.<\/span><\/p>\n

1)\u00a0\u00a0\u00a0 Rasgele Bir Say\u0131n\u0131n Sadece Ondokuzun Kat\u0131 Olarak \u0130fade Edilebilmesinin Olas\u0131l\u0131ksal De\u011feri<\/span><\/h1>\n

Rasgele se\u00e7ece\u011fimiz bir say\u0131n\u0131n 19\u2019un kat\u0131 olma olas\u0131l\u0131\u011f\u0131 her zaman (1\/19)dur. Se\u00e7imi (sadece 2 basamakl\u0131 say\u0131lar, sadece 3 basamakl\u0131 say\u0131lar gibi) daralt\u0131lm\u0131\u015f bir alanda yaparsak olas\u0131l\u0131k yine de (1\/19) civar\u0131nda kalmaya devam eder. Birbirinden ba\u011f\u0131ms\u0131z olaylar\u0131n birarada ger\u00e7ekle\u015fmesi durumunda olas\u0131l\u0131klar\u0131 \u00e7arp\u0131l\u0131r. Mesela (1\/19) olas\u0131l\u0131kl\u0131 birbirinden ba\u011f\u0131ms\u0131z n tane olay\u0131n birarada ger\u00e7ekle\u015fmesi olas\u0131l\u0131\u011f\u0131 (1\/19)n <\/sup>dir. Bu olas\u0131l\u0131\u011f\u0131n olay say\u0131s\u0131n\u0131n art\u0131\u015f\u0131ndan \u00e7ok \u00e7ok daha h\u0131zl\u0131 bir \u015fekilde artt\u0131\u011f\u0131n\u0131 g\u00f6sterir.\u00a0 Ondokuzda bir olas\u0131l\u0131kl\u0131 birka\u00e7 olay \u00fcst\u00fcste ger\u00e7ekle\u015fince olas\u0131l\u0131k teorisi bize, matematikle ilgili olmayanlara fazla anlam ifade etmeyecek, virg\u00fclden sonra bol s\u0131f\u0131rlar\u0131n ba\u015flad\u0131\u011f\u0131 say\u0131lar verir.<\/p>\n

Peki hangi seviyenin alt\u0131nda bir olas\u0131l\u0131\u011f\u0131 do\u011fal bir tesad\u00fcf olarak kabul edemeyiz? \u0130\u015fte matematik bu konuda bir standart \u00f6neremez. Burada matematik devreden \u00e7\u0131kar, vicdan\u0131m\u0131z ve sa\u011fduyumuz devreye girer. G\u00fcnl\u00fck hayatta hangi d\u00fczeydeki olas\u0131l\u0131klar\u0131 ciddiye al\u0131yor, hangilerini \u2018ger\u00e7ekle\u015fmez\u2019 diye ihmal ediyoruz? Bu tercih s\u0131n\u0131r\u0131n\u0131n herkes i\u00e7in ayn\u0131 olmas\u0131n\u0131 bekleyemeyiz. Mesela Bertrand Russell gibi septiklerde bu e\u015fik \u00e7ok\u00a0 daha d\u00fc\u015f\u00fck d\u00fczeyde olabilir. Ben insanlar\u0131n bu konuda belli bir e\u015fik de\u011feriyle de\u011fil fakat tutarl\u0131l\u0131klar\u0131yla sorumlu tutulaca\u011f\u0131na inan\u0131yorum<\/span>.<\/p>\n

Diyelim ki tan\u0131mad\u0131\u011f\u0131n\u0131z bir adamla barbut oynuyorsunuz. Siz 3 att\u0131n\u0131z, o 6… Ve paran\u0131z\u0131 ald\u0131. Siz 1 att\u0131n\u0131z, o yine 6… Ve yine paran\u0131z\u0131 ald\u0131. Siz ne atars\u0131n\u0131z at\u0131n o istisnas\u0131z hep 6 att\u0131 ve hep o kazand\u0131. Ka\u00e7\u0131nc\u0131 seferden sonra \u201cne kadar \u015fansl\u0131 adam\u201d <\/i>d\u00fc\u015f\u00fcncesi yerini \u201cacaba zar hileli mi?\u201d<\/i> sorusuna b\u0131rak\u0131r? Ve bu \u015f\u00fcphe ka\u00e7\u0131nc\u0131 seferden sonra doland\u0131r\u0131ld\u0131\u011f\u0131n\u0131z kanaatine d\u00f6n\u00fc\u015f\u00fcr?<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Tamam\u0131 19’un tam kat\u0131 gelen ba\u011f\u0131ms\u0131z verilerin say\u0131s\u0131<\/b><\/p>\n<\/td>\n

\n

Zarda
\nistisnas\u0131z s\u00fcrekli<\/span> 6 att\u0131\u011f\u0131 el say\u0131s\u0131<\/b><\/p>\n<\/td>\n

\n

Yaz\u0131-tura oyununda
\nistisnas\u0131z s\u00fcrekli<\/span> istedi\u011fini att\u0131\u011f\u0131 el say\u0131s\u0131<\/b><\/p>\n<\/td>\n<\/tr>\n

\n

2<\/b><\/p>\n<\/td>\n

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166<\/b><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Evet, matematik neye tesad\u00fcf, neye mucize diyece\u011finize kar\u0131\u015fmaz. Ama (1\/19) olas\u0131l\u0131kl\u0131 ka\u00e7 tane ba\u011f\u0131ms\u0131z olay\u0131n bir arada ger\u00e7ekle\u015fmesinin yukar\u0131daki barbut senaryosunun olas\u0131l\u0131ksal olarak ka\u00e7\u0131nc\u0131 eline denk geldi\u011fini g\u00f6sterebilir. Ya da benzer senaryonun ya\u015fand\u0131\u011f\u0131 bir yaz\u0131 tura oyununun… Ger\u00e7ek hayat\u0131n i\u00e7inden al\u0131nan senaryolarla teorik veriler aras\u0131nda ba\u011f kurmak, kendi tutarl\u0131l\u0131\u011f\u0131n\u0131 test etmek isteyen insanlar i\u00e7in g\u00fczel bir f\u0131rsat sunabilir.
\n19 d\u0131\u015f\u0131nda ba\u015fka bir asal say\u0131 kullan\u0131larak da elde edilen sonu\u00e7lar da b\u00f6yle d\u00fc\u015f\u00fck olas\u0131l\u0131klar verir mi? 19\u2019dan daha b\u00fcy\u00fck bir asal say\u0131 ile \u00e7al\u0131\u015f\u0131rsak buldu\u011fumuz her verinin etkisi daha fazla bile olur. Fakat \u00fcste \u00fcste, mesela 23\u2019\u00fcn katlar\u0131n\u0131 veren sistematik istatistiklerle kar\u015f\u0131la\u015fmak \u00e7ok \u00e7ok zor oldu\u011fundan Kuran\u2019da ya da ba\u015fka herhangi bir kitapta b\u00f6yle yap\u0131lar oldu\u011funu iddia eden \u00e7al\u0131\u015fmalar duymad\u0131m ve duyaca\u011f\u0131m\u0131 tahmin etmiyorum. 19\u2019dan k\u00fc\u00e7\u00fck say\u0131lar aras\u0131ndaysa (\u00f6zellikle 7 ve 2) b\u00f6yle \u00e7al\u0131\u015fmalar var. Ancak onlar\u0131n handikap\u0131 da olas\u0131l\u0131ksal olarak 19 ile ayn\u0131 etkiyi g\u00f6sterebilmeleri i\u00e7in \u00e7ok daha fazla say\u0131da ba\u011f\u0131ms\u0131z veri sunmalar\u0131 gerekiyor.\u00a0 Sonu\u00e7ta rasgele ald\u0131\u011f\u0131m\u0131z bir say\u0131n\u0131n 19\u2019un kat\u0131 \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 (1\/19) iken 7\u2019nin kat\u0131 \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 da (1\/7) olur ve tablomuz ona g\u00f6re \u015fekillenir. Tablomuza 7\u2019yi de eklersek:<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
\n

Tamam\u0131 7nin tam kat\u0131 gelen ba\u011f\u0131ms\u0131z verilerin say\u0131s\u0131<\/b><\/p>\n<\/td>\n

\n

Tamam\u0131 19’un tam kat\u0131 gelen ba\u011f\u0131ms\u0131z verilerin say\u0131s\u0131<\/b><\/p>\n<\/td>\n

\n

Zarda
\nistisnas\u0131z s\u00fcrekli
\n6 att\u0131\u011f\u0131 el say\u0131s\u0131<\/b><\/p>\n<\/td>\n

\n

Yaz\u0131-tura oyununda istisnas\u0131z s\u00fcrekli istedi\u011fini att\u0131\u011f\u0131 el say\u0131s\u0131<\/b><\/p>\n<\/td>\n<\/tr>\n

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3<\/b><\/p>\n<\/td>\n

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166<\/b><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

2) Ondokuzdan Vazge\u00e7meden Yan\u0131na Ba\u015fka Bir Say\u0131 Getirmek<\/span><\/strong><\/p>\n

Rasgele ald\u0131\u011f\u0131m\u0131z bir say\u0131n\u0131n 19\u2019un kat\u0131 \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 (1\/19) iken 7\u2019nin kat\u0131 \u00e7\u0131kma olas\u0131l\u0131\u011f\u0131 da (1\/7) olur demi\u015ftik. 19\u2019un kat\u0131 \u00e7\u0131kmayan bir say\u0131n\u0131n haliyle \u00e7arpanlar\u0131 da 19\u2019un kat\u0131 \u00e7\u0131kmaz. Peki 19\u2019un kat\u0131 \u00e7\u0131kmayan bir say\u0131y\u0131 biri 19\u2019un kat\u0131 di\u011feri se\u00e7ece\u011fimiz karde\u015f bir say\u0131n\u0131n (mesela 7 olsun) kat\u0131 olacak \u015fekilde iki toplanana ay\u0131rsak ve bu yolla tamam\u0131 19 ve 7\u2019nin katlar\u0131n\u0131n toplam\u0131 edecek olan bir s\u00fcr\u00fc ba\u011f\u0131ms\u0131z verilerden olu\u015fan bir yap\u0131 elde etsek; bu yap\u0131n\u0131n olas\u0131l\u0131ksal de\u011feri ne olurdu? Tutarl\u0131 bir tepki vermek ad\u0131na bizi ne kadar \u015fa\u015f\u0131rtmal\u0131yd\u0131?<\/p>\n

Do\u011fru cevap hi\u00e7. Sadece 19\u2019un, sadece 7\u2019nin, sadece x\u2019in katlar\u0131n\u0131 aramak \u00fczerine in\u015fa edilen bir yap\u0131n\u0131n heyecan yaratmas\u0131 ortaya \u00e7\u0131kacak verilere g\u00f6re ihtimal dahilindedir. Ama tek bir say\u0131 yerine bir say\u0131 ikilisi kullanmak sistemi daha enteresan de\u011fil, s\u0131radan hale getirir. Matematiksel olarak aralar\u0131nda asal<\/b>, yani hi\u00e7bir ortak b\u00f6leni olmayan iki say\u0131n\u0131n katlar\u0131 toplam\u0131<\/b>n\u0131 aramak-bulmak de\u011fil, ona rastlamamak mucize olur. 19 bir asal say\u0131 olmasayd\u0131, yan\u0131na bir say\u0131 getirip ikili olu\u015fturmaktan belki bahsedilebilirdi. Ama 19 asal say\u0131 oldu\u011fu i\u00e7in yan\u0131na getirelecek her say\u0131 ile illa ki aralar\u0131nda asal olacaklard\u0131r. Bu da linear kombinasyonlar\u0131 aras\u0131nda 1\u2019in oldu\u011fu anlam\u0131na gelir. 1\u2019i elde eden de her say\u0131y\u0131 elde edebilir.<\/p>\n

\u0130\u015fin teorik k\u0131sm\u0131na fazla girmeden herhangi bir say\u0131n\u0131n 7 ve 19\u2019un katlar\u0131 cinsinden ifade edebilece\u011fimiz bir y\u00f6ntem sunay\u0131m.
\nHerhangibir say\u0131 alal\u0131m. Mesela 6348 olsun. J<\/p>\n

6348 19\u2019a b\u00f6l\u00fcnm\u00fcyor. 7\u2019ye de b\u00f6l\u00fcnm\u00fcyor… Hi\u00e7 \u00fcz\u00fclm\u00fcyoruz. Birazdan bu say\u0131dan 19\u2019un tam kat\u0131 olan \u00f6yle bir par\u00e7a kopartaca\u011f\u0131z ki, geriye kalan k\u0131s\u0131m 7\u2019ye kalans\u0131z b\u00f6l\u00fcnecek. \u015eimdi sakince 7\u2019ye b\u00f6l\u00fcm\u00fcnden ka\u00e7 artt\u0131\u011f\u0131na bak\u0131yoruz:<\/p>\n

6348\/7 = (906 x 7) + 9<\/p>\n

Sonra hemen 19\u2019un tamkat\u0131 olan say\u0131lar aras\u0131ndan 7\u2019ye b\u00f6l\u00fcm\u00fc 6348 ile ayn\u0131 kalan\u0131 verecek bir tanesini se\u00e7iyoruz. B\u00f6yle ka\u00e7 tane say\u0131 var se\u00e7ebilece\u011fimiz? Sonsuz…\u00a0Ben a\u015fa\u011f\u0131da birka\u00e7 tanesini listeledim.<\/p>\n\n\n\n\n\n\n\n\n\n
\n

19’un tamkat\u0131 olan say\u0131lar\u0131n 7’ye b\u00f6l\u00fcmden kalanlar<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n

\n

133<\/p>\n<\/td>\n

\n

266<\/p>\n<\/td>\n

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399<\/p>\n<\/td>\n

\n

532<\/p>\n<\/td>\n

\n

665<\/p>\n<\/td>\n

\n

798<\/p>\n<\/td>\n

\n

931<\/p>\n<\/td>\n

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1064<\/p>\n<\/td>\n

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1197<\/p>\n<\/td>\n

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1330<\/p>\n<\/td>\n

…<\/td>\n<\/tr>\n
\n

1<\/p>\n<\/td>\n

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57<\/p>\n<\/td>\n

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323<\/p>\n<\/td>\n

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589<\/p>\n<\/td>\n

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722<\/p>\n<\/td>\n

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855<\/p>\n<\/td>\n

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988<\/p>\n<\/td>\n

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1121<\/p>\n<\/td>\n

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1254<\/p>\n<\/td>\n

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1387<\/p>\n<\/td>\n

…<\/td>\n<\/tr>\n
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2<\/p>\n<\/td>\n

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114<\/p>\n<\/td>\n

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380<\/p>\n<\/td>\n

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646<\/p>\n<\/td>\n

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779<\/p>\n<\/td>\n

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1178<\/p>\n<\/td>\n

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1311<\/p>\n<\/td>\n

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…<\/td>\n<\/tr>\n
\n

3<\/p>\n<\/td>\n

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1368<\/p>\n<\/td>\n

…<\/td>\n<\/tr>\n
\n

4<\/p>\n<\/td>\n

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95<\/p>\n<\/td>\n

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1292<\/p>\n<\/td>\n

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1425<\/p>\n<\/td>\n

…<\/td>\n<\/tr>\n
\n

5<\/p>\n<\/td>\n

\n

19<\/p>\n<\/td>\n

\n

152<\/p>\n<\/td>\n

\n

285<\/p>\n<\/td>\n

\n

418<\/p>\n<\/td>\n

\n

551<\/p>\n<\/td>\n

\n

684<\/p>\n<\/td>\n

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817<\/p>\n<\/td>\n

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950<\/p>\n<\/td>\n

\n

1083<\/p>\n<\/td>\n

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1216<\/p>\n<\/td>\n

\n

1349<\/p>\n<\/td>\n

…<\/td>\n<\/tr>\n
\n

6<\/b><\/p>\n<\/td>\n

\n

76<\/b><\/p>\n<\/td>\n

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209<\/b><\/p>\n<\/td>\n

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342<\/b><\/p>\n<\/td>\n

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475<\/b><\/p>\n<\/td>\n

\n

608<\/b><\/p>\n<\/td>\n

\n

741<\/b><\/p>\n<\/td>\n

\n

874<\/b><\/p>\n<\/td>\n

\n

1007<\/b><\/p>\n<\/td>\n

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1140<\/b><\/p>\n<\/td>\n

\n

1273<\/b><\/p>\n<\/td>\n

\n

1406<\/b><\/p>\n<\/td>\n

…<\/b><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

6348, 7\u2019nin katlar\u0131ndan 6 artt\u0131rd\u0131\u011f\u0131 i\u00e7in 7\u2019nin katlar\u0131ndan 6 artt\u0131racak bir say\u0131y\u0131 se\u00e7iyoruz.<\/p>\n

76, 209, 342, 475,…<\/p>\n

133\u2019er 133\u2019er (133=19×7) arttt\u0131\u011f\u0131n\u0131 farkedebilece\u011finiz bu listedeki say\u0131lardan diledi\u011fimizi se\u00e7iyoruz.
\nHaydi 1140 olsun. Geriye ka\u00e7 kald\u0131?<\/p>\n

6348-1140=5208<\/p>\n

B\u00f6ylece<\/p>\n

6348 = 1140 + 5208<\/p>\n

6348 = (60 x 19) + (744 x 7)<\/p>\n

\u0130ki tarafa daha dengeli da\u011f\u0131tmak istersek, mesela:<\/p>\n

6348 = 3135 + 3213 say\u0131lar\u0131n\u0131 se\u00e7ebiliriz. B\u00f6ylece:<\/p>\n

6348 = (165 x 19) + (459 x 7) diye de yazabiliriz.<\/p>\n

Sadece 6348\u2019i de\u011fil istisnas\u0131z b\u00fct\u00fcn say\u0131lar\u0131 19 ve 7\u2019nin katlar\u0131 cinsinden elde edebiliriz.<\/p>\n

19\u2019a tam b\u00f6l\u00fcnmeyen, (geride kalan b\u0131rakan) her say\u0131 ya 7\u2019ye tam b\u00f6l\u00fcn\u00fcr ya da 7 ile 19\u2019un katlar\u0131 cinsinden ifade edilebilir. <\/b>Hem de sonsuz farkl\u0131 \u015fekilde…
\nBu durum k\u00fc\u00e7\u00fck say\u0131lar i\u00e7in de ge\u00e7erlidir. Ama k\u00fc\u00e7\u00fck say\u0131lar iki b\u00fcy\u00fck say\u0131n\u0131n pozitif katlar\u0131n\u0131n toplam\u0131 olarak yaz\u0131lamayaca\u011f\u0131ndan negatif katlar\u0131 kullanmak yani \u00e7\u0131karma yapmak gerekecektir.
\nMesela:<\/p>\n

1 = 57 \u2013 56<\/p>\n

1= (3 x19) \u2013 (8 x 7) gibi…<\/p>\n

114\u2019ten k\u00fc\u00e7\u00fck say\u0131lar\u0131n\u0131 yar\u0131s\u0131na yak\u0131n\u0131 i\u00e7in \u00e7\u0131karma yapmak bir ihtiya\u00e7. \u0130\u015fin i\u00e7ine \u00e7\u0131karma ya da negatif say\u0131lar girmesin istiyorsak:<\/p>\n

\u201c114\u2019den b\u00fcy\u00fck, 19\u2019a tam b\u00f6l\u00fcnmeyen her say\u0131 ya 7\u2019ye tam b\u00f6l\u00fcn\u00fcr ya da 7 ile 19\u2019un pozitif katlar\u0131 cinsinden ifade edilebilir.\u201d <\/b>de diyebiliriz.<\/b><\/p>\n

\u00d6zet olarak,<\/b><\/p>\n

    \n<\/ol>\n
      \n
    1. Say\u0131lar\u0131n az bir k\u0131sm\u0131<\/span> (%5\u2019i) 19\u2019a tam b\u00f6l\u00fcn\u00fcr (19\u2019un tam kat\u0131d\u0131r).<\/li>\n
    2. Geri kalan t\u00fcm<\/span> say\u0131lar ya 7\u2019nin tam kat\u0131d\u0131r, ya da 19 ve 7\u2019nin tamkatlar\u0131 olacak iki par\u00e7aya ayr\u0131labilir.<\/li>\n
    3. 19\u2019un geride b\u0131rakt\u0131klar\u0131n\u0131 7 ile i\u00e7eriye alma \u00e7abalar\u0131<\/span> olas\u0131l\u0131ksal a\u00e7\u0131dan hi\u00e7bir\u015fey ifade etmez.<\/li>\n<\/ol>\n

      7ar1k<\/p>\n

      TARTI\u015eMALAR:<\/p>\n

      \u0130MRAN AKDEM\u0130R:<\/b> Tar\u0131k Arabac\u0131, sen var ya sen… 7 ve 19\u2019un sistemde her tabloda ayn\u0131 anda sonu\u00e7land\u0131\u011f\u0131nda bunun 1\/133 ile 1\/67 aras\u0131nda oldu\u011funu ve 19 olas\u0131l\u0131\u011f\u0131na nazaran bir u\u00e7urum kadar fark oldu\u011funu bilenlerden olmana ra\u011fmen, basit bir mod\u00fcler aritmetik i\u015flemiyle kendini ve etraftakileri yan\u0131lt\u0131yorsun. Makaledeki o en son tabloya, yani 6348\u2019e +1 veya -1 say\u0131s\u0131 i\u015fleme girdi\u011finde, tablo tamamen i\u015flevsiz olur. Bu tablonun aptalca oldu\u011funu sen \u00e7ok iyi biliyorsun!<\/p>\n

      TARIK ARABACI:<\/b> +1 veya -1’leri i\u015fleme dahil edersek 7’nin kat\u0131 olman\u0131n olas\u0131l\u0131k de\u011feri 1\/7 de\u011fil 3\/7 olur. Yakla\u015f\u0131k yar\u0131 yar\u0131ya yani…<\/p>\n

      7&19’lu tablolardan birini “1 tane 19’lu istatistik”le kar\u015f\u0131la\u015ft\u0131r\u0131rsak evet hakl\u0131s\u0131n o tablolar daha de\u011ferli olur. Ama mesela 4 tane 19’lu bir tabloyla kar\u015f\u0131la\u015ft\u0131r\u0131rsak bu kez aralar\u0131nda ger\u00e7ekten u\u00e7urum olur. Ama 19 lehine…<\/p>\n

      Hay\u0131r 6348 sadece bir \u00f6rnek. 6348 yerine 6347 ya da 6349 (hatta cep telefonu numaram bile) olsa o tablo ve algoritmas\u0131 hala ge\u00e7erli. O tablodaki y\u00f6ntemi 6347’ye de uygulad\u0131m. 48 farkl\u0131 ikili buldum. \u0130kililerden biri 7’nin kat\u0131, di\u011feri 19’un… Hepsinin de toplam\u0131 6347. Hepsinin toplam\u0131 Se\u00e7, be\u011fen, al. (6349 i\u00e7in de 47 tane bulabilirim)<\/p>\n

      19 + 6328<\/p>\n

      77 + 6270<\/p>\n

      152 + 6195<\/p>\n

      210 + 6137<\/p>\n

      285 + 6062<\/p>\n

      343 + 6004<\/p>\n

      418 + 5929<\/p>\n

      476 + 5871<\/p>\n

      551 + 5796<\/p>\n

      609 + 5738<\/p>\n

      684 + 5663<\/p>\n

      742 + 5605<\/p>\n

      817 + 5530<\/p>\n

      875 + 5472<\/p>\n

      950 + 5397<\/p>\n

      1008 + 5339<\/p>\n

      1083 + 5264<\/p>\n

      1141 + 5206<\/p>\n

      1216 + 5131<\/p>\n

      1274 + 5073<\/p>\n

      1349 + 4998<\/p>\n

      1407 + 4940<\/p>\n

      1482 + 4865<\/p>\n

      1540 + 4807<\/p>\n

      1615 + 4732<\/p>\n

      1673 + 4674<\/p>\n

      1748 + 4599<\/p>\n

      1806 + 4541<\/p>\n

      1881 + 4466<\/p>\n

      1939 + 4408<\/p>\n

      2014 + 4333<\/p>\n

      2072 + 4275<\/p>\n

      2147 + 4200<\/p>\n

      2205 + 4142<\/p>\n

      2280 + 4067<\/p>\n

      2338 + 4009<\/p>\n

      2413 + 3934<\/p>\n

      2471 + 3876<\/p>\n

      2546 + 3801<\/p>\n

      2604 + 3743<\/p>\n

      2679 + 3668<\/p>\n

      2737 + 3610<\/p>\n

      2812 + 3535<\/p>\n

      2870 + 3477<\/p>\n

      2945 + 3402<\/p>\n

      3003 + 3344<\/p>\n

      3078 + 3269<\/p>\n

      3136 + 3211<\/p>\n

       <\/p>\n

      Toplamlar\u0131 cep telefonu numaran\u0131 veren biri 7 di\u011feri 19’un kat\u0131 olacak 40 milyondan fazla ikili var. Hatta bu ikililerden biri sabit ev telefonu numaran \u00e7\u0131ksa bile \u015fa\u015f\u0131rmamal\u0131.\u00a0 \u00c7\u00fcnk\u00fc 14 milyon Telekom abonesinin 200bininden fazlas\u0131n\u0131n ev-cep numaralar\u0131 aras\u0131nda b\u00f6yle 7 & 19’lu bir ba\u011flant\u0131 var.<\/p>\n","protected":false},"excerpt":{"rendered":"

      \u00d6zet olarak,<\/p>\n

      Say\u0131lar\u0131n az bir k\u0131sm\u0131 (%5\u2019i) 19\u2019a tam b\u00f6l\u00fcn\u00fcr (19\u2019un tam kat\u0131d\u0131r).
      \nGeri kalan t\u00fcm say\u0131lar ya 7\u2019nin tam kat\u0131d\u0131r, ya da 19 ve 7\u2019nin tamkatlar\u0131 olacak iki par\u00e7aya ayr\u0131labilir.
      \n19\u2019un geride b\u0131rakt\u0131klar\u0131n\u0131 7 ile i\u00e7eriye alma \u00e7abalar\u0131 olas\u0131l\u0131ksal a\u00e7\u0131dan hi\u00e7bir\u015fey ifade etmez.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[161,1542],"tags":[1545,1544,1543],"class_list":["post-6102","post","type-post","status-publish","format-standard","hentry","category-19-ayet","category-tarik-arabaci","tag-ikiz-kod","tag-ikizkod","tag-imran-akdemir"],"acf":[],"_links":{"self":[{"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/posts\/6102","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/comments?post=6102"}],"version-history":[{"count":1,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/posts\/6102\/revisions"}],"predecessor-version":[{"id":6180,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/posts\/6102\/revisions\/6180"}],"wp:attachment":[{"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/media?parent=6102"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/categories?post=6102"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/19.org\/tr\/wp-json\/wp\/v2\/tags?post=6102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}