{"id":19329,"date":"2017-11-01T12:45:32","date_gmt":"2017-11-01T09:45:32","guid":{"rendered":"http:\/\/109.232.216.219\/~bilimvegelecek\/?p=19329"},"modified":"2018-02-13T12:49:28","modified_gmt":"2018-02-13T09:49:28","slug":"1-saniyede-matematik","status":"publish","type":"post","link":"https:\/\/bilimvegelecek.com.tr\/index.php\/2017\/11\/01\/1-saniyede-matematik","title":{"rendered":"1 saniyede matematik!"},"content":{"rendered":"

Bir k\u00fcmeste 100 civciv halka \u015feklinde ve huzur i\u00e7inde otururlarken aniden birbirlerini gagalamaya ba\u015flarlar. Her bir civciv solundaki veya sa\u011f\u0131ndaki\u00a0 civcivi gagalad\u0131\u011f\u0131na g\u00f6re ka\u00e7 civcivin gagalanmam\u0131\u015f olmas\u0131 beklenir?<\/em><\/p>\n

Bu soru ge\u00e7ti\u011fimiz May\u0131s ay\u0131nda Amerika\u2019da d\u00fczenlenen bir matematik yar\u0131\u015fmas\u0131nda son soru olarak sorulmu\u015f. 13 ya\u015f\u0131ndaki 7\u2019inci s\u0131n\u0131f \u00f6\u011frencisi Luke Robitaille, sunucunun soruyu okumas\u0131n\u0131 bitirmesini beklemeden 1 saniyeden k\u0131sa bir s\u00fcrede yan\u0131t vererek salondakilerin \u015fa\u015fk\u0131n bak\u0131\u015flar\u0131 ve alk\u0131\u015flar aras\u0131nda \u015fampiyonlu\u011funu ilan etmi\u015f.<\/p>\n

Merak ettim yar\u0131\u015fman\u0131n videosunu izledim, me\u011ferse Luke\u2019un h\u0131z yapt\u0131\u011f\u0131 ilk soru bu de\u011filmi\u015f. Rakibinden \u00f6nce butona basarak bir\u00e7ok soruyu 1 saniyede yan\u0131tlamay\u0131 ba\u015farm\u0131\u015f. Bu sorulardan bir ka\u00e7\u0131n\u0131 bu yaz\u0131da bulabilirsiniz ve eminim ki siz de o sorulara verilen saniyelik cevaplar kar\u015f\u0131s\u0131nda \u00e7ok \u015fa\u015f\u0131racaks\u0131n\u0131z. Asl\u0131nda b\u00f6ylesine bir h\u0131z matematik yapman\u0131n do\u011fas\u0131na ayk\u0131r\u0131 say\u0131labilir, \u00e7\u00fcnk\u00fc matematikle u\u011fra\u015f\u0131yor olmak zamandan ba\u011f\u0131ms\u0131z olarak anlamay\u0131, ara\u015ft\u0131rmay\u0131, ke\u015ffetmeyi gerektirir. H\u0131z gerekmez! Ama \u00f6te yandan bu t\u00fcr yar\u0131\u015fmalarda \u00f6ne \u00e7\u0131kan isimlerin yetenekleri ve \u00e7al\u0131\u015fkanl\u0131klar\u0131 su g\u00f6t\u00fcrmez bir ger\u00e7ek. Umar\u0131m ki, Luke Robitalle ileride matematik\u00e7i olur ve \u00e7al\u0131\u015fmalar\u0131yla ad\u0131ndan s\u00f6z ettirir.<\/p>\n

Yar\u0131\u015fma hakk\u0131nda edindi\u011fim bilgiler \u015f\u00f6yle: Alt\u0131, yedi ve sekizinci s\u0131n\u0131f \u00f6\u011frencilerine a\u00e7\u0131k olan yar\u0131\u015fmada \u00f6nce 224 \u00f6\u011frenciye yaz\u0131l\u0131 bir test uygulan\u0131yor. En y\u00fcksek puan\u0131 alan 12 \u00f6\u011frenci ertesi g\u00fcn sorular\u0131n s\u00f6zl\u00fc olarak okundu\u011fu 1000 ki\u015filik bir salonda yar\u0131\u015f\u0131yor. Her bir soruya en \u00e7ok 45 saniye s\u00fcre veriliyor. Final b\u00f6l\u00fcm\u00fcnde 4 soruyu do\u011fru yan\u0131tlayan ilk \u00f6\u011frenci yar\u0131\u015fmay\u0131 kazan\u0131yor. \u00a0\u015eampiyon olan 20.000 Dolarl\u0131k \u00fcniversite bursunu ve uzay kamp\u0131nda konaklama hakk\u0131n\u0131 kazan\u0131yor.<\/p>\n

\u015eampiyon Luke, \u201ccivciv problemini\u201d yan\u0131tlamadan \u00f6nce rakibiyle e\u015fit say\u0131da soruyu do\u011fru yan\u0131tlam\u0131\u015f, yani bu sorudan \u00f6nce durum 3-3. \u201cCivciv sorusunu\u201d do\u011fru yan\u0131tlayarak 4-3\u2019l\u00fck skorla birincili\u011fi kazanm\u0131\u015f. Bu sorudan \u00f6nceki soruyu da 1 saniyede yan\u0131tl\u0131yor. \u0130\u015fte o soru:<\/p>\n

Alt\u0131 basamakl\u0131 pozitif tamsay\u0131lar\u0131n ka\u00e7\u0131 1000 ile b\u00f6l\u00fcnebilirken 400 ile b\u00f6l\u00fcnemez?<\/em><\/p>\n

Luke\u2019un sunucunun soruyu okumas\u0131n\u0131 bitirmesini beklemeden 1 saniyeden k\u0131sa bir s\u00fcrede yan\u0131tlad\u0131\u011f\u0131 bir soru daha:<\/p>\n

4x<\/sup> + 5x<\/sup> = 6 x<\/sup> denklemini sa\u011flayan x say\u0131s\u0131ndan k\u00fc\u00e7\u00fck en b\u00fcy\u00fck tamsay\u0131 ka\u00e7t\u0131r?<\/em><\/p>\n

\u201cBe\u015fer \u015fa\u015far Luke da \u015fa\u015far\u201d diyelim ve bu kez onun yanl\u0131\u015f rakibinin do\u011fru yan\u0131tlad\u0131\u011f\u0131 bir soru yazal\u0131m:<\/p>\n

Ian bir merdiveni \u00e7\u0131karken her defas\u0131nda rastgele olarak 1 veya 2 veya 3 ad\u0131m atmaya karar veriyor. \u0130lk basamaktan ad\u0131m atmaya ba\u015flayan Ian\u2019\u0131n 4\u2019\u00fcnc\u00fc basama\u011fa basma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r?<\/em><\/p>\n

Bu yar\u0131\u015fmada sorulan di\u011fer sorular\u0131 merak eden okur Kaynak 2\u2019de adresini verdi\u011fim videoyu izleyebilir.<\/p>\n

Kendini denemek isteyen okurlar olabilir diye sorular\u0131n \u00e7\u00f6z\u00fcmlerini \u00f6zellikle vermedim. Civciv sorusunu \u00e7\u00f6zebildiniz mi? Ka\u00e7 saniyede? A\u00e7\u0131k\u00e7as\u0131 bu sat\u0131rlar\u0131n yazar\u0131 i\u00e7in maalesef dakikalar gerekti. Merakl\u0131 okurlar i\u00e7in yukar\u0131daki sorular\u0131n \u00e7\u00f6z\u00fcmlerini veriyorum.<\/p>\n

\u00c7\u00f6z\u00fcmler<\/strong><\/p>\n

Civciv problemi:<\/strong> Herhangi bir civcivin gagalanmamas\u0131 olas\u0131l\u0131\u011f\u0131n\u0131 bulal\u0131m. Y\u00fczleri ayn\u0131 y\u00f6ne bakan A, B, C<\/em> olarak s\u0131ralanm\u0131\u015f \u00fc\u00e7 civcivden ortadaki civciv olan B<\/em>\u2019nin gagalanmamas\u0131 i\u00e7in A<\/em>\u2019n\u0131n sa\u011f\u0131ndaki civcivi, C<\/em>\u2019ninse solundakini gagalamas\u0131 gerekiyor. A<\/em>\u2019n\u0131n sa\u011f\u0131ndaki civcivi gagalamas\u0131 olas\u0131l\u0131\u011f\u0131 1\/2, C<\/em>\u2019nin solundakini gagalamas\u0131 olas\u0131l\u0131\u011f\u0131 da 1\/2 oldu\u011fundan bu iki olay\u0131n birlikte ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131<\/p>\n

1\/2 x 1\/2 = 1\/4<\/p>\n

olur. Bu sonu\u00e7 bir civcivin (B<\/em>\u2019nin) gagalanmamas\u0131 olas\u0131l\u0131\u011f\u0131d\u0131r. 100 civciv i\u00e7inde gagalanmam\u0131\u015f civcivlerin muhtemel say\u0131s\u0131, yani beklenti de\u011feri<\/em><\/p>\n

100 x 1\/4 =25<\/p>\n

olur. Burada s\u00f6z\u00fc edilen beklenti de\u011feri <\/em>olas\u0131l\u0131k teorisinde yer alan bir kavramd\u0131r. Bir olay\u0131n beklenti de\u011feri (matematiksel umudu<\/em> da diyebiliriz) kabaca \u015f\u00f6yle hesaplan\u0131r: O olay\u0131n ger\u00e7ekle\u015fme olas\u0131l\u0131\u011f\u0131yla, olay\u0131 ifade eden miktar\u0131n say\u0131sal de\u011ferinin \u00e7arp\u0131m\u0131d\u0131r.<\/p>\n

Bu problemin \u00e7\u00f6z\u00fcm\u00fcn\u00fc \u015f\u00f6yle de d\u00fc\u015f\u00fcnebiliriz: Her bir civciv i\u00e7in d\u00f6rt se\u00e7enek vard\u0131r: 1) Sa\u011f\u0131ndan gagalanabilir. 2) Solundan gagalanabilir. 3) Hem solundan hem sa\u011f\u0131ndan gagalanabilir. 4) Hi\u00e7 gagalanmayabilir. Dolay\u0131s\u0131yla 100 civciv i\u00e7inde\u00a0 hi\u00e7\u00a0gagalanmayan civcivlerin beklenti de\u011feri 100\/4 =25 olur.<\/p>\n

Alt\u0131 basamakl\u0131 say\u0131lar problemi<\/strong>: 1000 ile b\u00f6l\u00fcnebilen alt\u0131 basamakl\u0131 en k\u00fc\u00e7\u00fck say\u0131 100.000\u2019dir, ama bu say\u0131 400\u2019e b\u00f6l\u00fcn\u00fcr. 101.000, 1000\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fc halde 400\u2019e b\u00f6l\u00fcnmez, \u00e7\u00fcnk\u00fc \u00f6nce 100\u2019e b\u00f6ld\u00fc\u011f\u00fcm\u00fczde ard\u0131ndan 4\u2019e b\u00f6lemeyiz. Bu \u00f6zelli\u011fe sahip say\u0131lar 101, 103,\u2026999 gibi \u00fc\u00e7 basamakl\u0131 t\u00fcm tek say\u0131lar\u0131n 1000 kat\u0131 olan alt\u0131 basamakl\u0131 say\u0131lard\u0131r. O halde 100 ile 1000 aras\u0131ndaki tek say\u0131lar\u0131n ka\u00e7 tane oldu\u011funu bulmal\u0131y\u0131z: (999-101)\/2+1=450.<\/p>\n

Denklem: <\/strong>4x<\/sup> + 5x<\/sup> = 6 x<\/sup> e\u015fitli\u011finde x=2 al\u0131n\u0131rsa 41>36, x=3 al\u0131n\u0131rsa 189<216 olur. O halde x say\u0131s\u0131 2 ile 3 aras\u0131ndad\u0131r. Sorunun cevab\u0131 2 olur.<\/p>\n

Olas\u0131l\u0131k sorusu:<\/strong> \u00d6nce Ian\u2019\u0131n 1\u2019inci basama\u011fa basma olas\u0131l\u0131\u011f\u0131n\u0131 yazal\u0131m: 1\/3, \u00e7\u00fcnk\u00fc ba\u015flang\u0131\u00e7tan itibaren ya 1, ya 2, ya da 3 basamak kadar ad\u0131m at\u0131yor.<\/p>\n

2\u2019inci basama\u011fa basma olas\u0131l\u0131\u011f\u0131n\u0131 ise \u015f\u00f6yle hesaplayabiliriz:\u00a0 1\u2019inci basama\u011fa bast\u0131ktan sonra ya 2\u2019ye basar ya da do\u011frudan 2\u2019ye basar. Bu olas\u0131l\u0131k,<\/p>\n

1\/3 x 1\/3 + 1\/3 = 4\/9<\/p>\n

olur.<\/p>\n

3\u2019\u00fcnc\u00fc basama\u011fa basma olas\u0131l\u0131\u011f\u0131n\u0131 da benzer \u015fekilde hesaplayabiliriz: \u00d6nce 1\u2019e sonra 3\u2019e veya \u00f6nce 2\u2019ye sonra 3\u2019e veya do\u011frudan 3\u2019e basar. Bu durumda<\/p>\n

1\/3 x 1\/3 + 1\/4 x 1\/3 + 1\/3 = 16\/27<\/p>\n

bulunur.<\/p>\n

4\u2019\u00fcnc\u00fc basama\u011fa basma olas\u0131l\u0131\u011f\u0131 ise yine \u00f6nce 1\u2019e sonra 4\u2019e veya \u00f6nce 2\u2019ye sonra 4\u2019e veya \u00f6nce 3\u2019e sonra 4\u2019e basma olas\u0131l\u0131klar\u0131n\u0131n toplam\u0131d\u0131r. Bu durumda soruda istenen olas\u0131l\u0131k<\/p>\n

1\/3 x 1\/3 + 4\/9 + 1\/3 + 16\/27 x 1\/3 = 37\/81<\/p>\n

olarak bulunur. (Bu olas\u0131l\u0131k problemi Luke\u2019un en fazla zaman harcad\u0131\u011f\u0131 sorulardan biri olmu\u015f ve yanl\u0131\u015f yan\u0131t vererek cevab\u0131n 41\/81 oldu\u011funu s\u00f6ylemi\u015f. Rakibi ise ondan hemen sonra s\u00f6z alarak do\u011fru yan\u0131t\u0131 s\u00f6ylemi\u015f ve skoru 2-2 yapm\u0131\u015f.)<\/p>\n

Amerika\u2019da \u00f6zel bir \u015firket taraf\u0131ndan d\u00fczenlenmi\u015f olan bu yar\u0131\u015fma New York Times\u2019a haber olacak kadar ilgi \u00e7ekmi\u015f. Akl\u0131ma, 16\u2019\u0131nc\u0131 y\u00fczy\u0131lda Avrupa\u2019da d\u00fczenlenen halka a\u00e7\u0131k, y\u00fczlerce ki\u015finin izledi\u011fi matematik yar\u0131\u015fmalar\u0131 geliyor. Bu yar\u0131\u015fmalar\u0131n birinde \u00fc\u00e7\u00fcnc\u00fc dereceden denklem yar\u0131\u015fmas\u0131n\u0131 kazanan matematik\u00e7i Tartaglia\u2019n\u0131n bu denklemlerin genel \u00e7\u00f6z\u00fcm\u00fcn\u00fc ke\u015ffetti\u011fini an\u0131ms\u0131yorum. G\u00fcn\u00fcm\u00fczde de Perelman gibi bir\u00e7ok \u00fcnl\u00fc matematik\u00e7inin orta \u00f6\u011frenim y\u0131llar\u0131nda kat\u0131ld\u0131klar\u0131 yar\u0131\u015fma ve olimpiyatlardaki ba\u015far\u0131lar\u0131n\u0131 hat\u0131rl\u0131yorum. Ve umut ediyorum\u2026 Bir g\u00fcn \u00fclkemizde de b\u00f6ylesi matematik yar\u0131\u015fmalar\u0131 ilgiyle kar\u015f\u0131lanacak ve gelece\u011fin matematik\u00e7ilerinin ke\u015ffedilmesine vesile olacak.<\/p>\n

Te\u015fekk\u00fcr.<\/strong> Yaz\u0131da s\u00f6z\u00fc edilen yar\u0131\u015fmadan ve \u201ccivciv problem\u201dinden beni haberdar eden Do\u011fucan G\u00fcnc\u00fc\u2019ye te\u015fekk\u00fcr ederim.<\/p>\n

Kaynaklar<\/strong><\/p>\n

– https:\/\/www.nytimes.com\/…\/math-counts-national-competiti<\/p>\n

– http:\/\/hmongbuy.net\/video\/vFTeN17Z4rc<\/p>\n","protected":false},"excerpt":{"rendered":"

Bir k\u00fcmeste 100 civciv halka \u015feklinde ve huzur i\u00e7inde otururlarken aniden birbirlerini gagalamaya ba\u015flarlar. Her bir civciv solundaki veya sa\u011f\u0131ndaki\u00a0 civcivi gagalad\u0131\u011f\u0131na g\u00f6re ka\u00e7 civcivin gagalanmam\u0131\u015f olmas\u0131 beklenir? Bu soru ge\u00e7ti\u011fimiz May\u0131s ay\u0131nda Amerika\u2019da d\u00fczenlenen bir matematik yar\u0131\u015fmas\u0131nda son soru olarak sorulmu\u015f. 13 ya\u015f\u0131ndaki 7\u2019inci s\u0131n\u0131f \u00f6\u011frencisi Luke Robitaille, sunucunun soruyu okumas\u0131n\u0131 bitirmesini beklemeden 1 […]<\/p>\n","protected":false},"author":375,"featured_media":19330,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1131,25,514],"tags":[208,1572,1530],"class_list":["post-19329","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-165-sayi","category-matematik","category-matematik-sohbetleri","tag-matematik","tag-problemler","tag-sayilar"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/19329","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/users\/375"}],"replies":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/comments?post=19329"}],"version-history":[{"count":0,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/19329\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media\/19330"}],"wp:attachment":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media?parent=19329"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/categories?post=19329"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/tags?post=19329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}