– Peki, bo\u015fk\u00fcmeye g\u00fcnl\u00fck hayattan ne gibi \u00f6rnekler verilebilir?<\/p>\n
– Dostoyevski\u2019nin Karamazov Karde\u015fler <\/em>roman\u0131n\u0131 be\u015f dakikada okuyabilen okurlar k\u00fcmesi, y\u00fcz metreyi iki saniyede ko\u015fabilen ki\u015filer k\u00fcmesi gibi bir\u00e7ok \u00f6rnek sayabiliriz. Ama sadece bir tane bo\u015fk\u00fcme vard\u0131r.Tabii bu bir \u00f6nerme ve sezgilerimiz do\u011fru oldu\u011funu s\u00f6yl\u00fcyor. Ama kan\u0131tlayabiliriz. Nas\u0131l?<\/p>\n – \u00c7eli\u015fki y\u00f6ntemiyle kan\u0131tlasak, yani hi\u00e7 eleman\u0131 olmayan iki tane k\u00fcme oldu\u011funu varsayal\u0131m.<\/p>\n – \u00c7ok iyi, sonra\u2026<\/p>\n – Hi\u00e7 eleman\u0131 olmayan bu iki k\u00fcmeye A ve B diyelim. A=B e\u015fitli\u011fini kan\u0131tlarsak sadece bir tane bo\u015fk\u00fcme var demektir. \u015eimdi, A ve B k\u00fcmelerinin birbirine e\u015fit olmad\u0131\u011f\u0131n\u0131 varsayal\u0131m. Bu durumda ikisinden birinde, di\u011ferinde olmayan, en az bir eleman bulunur; \u00e7\u00fcnk\u00fc aksi halde bu k\u00fcmeler birbirine e\u015fit olurdu. Oysa bu k\u00fcmelerin hi\u00e7 eleman\u0131 yok! Bu y\u00fczden ikisinden birinde, di\u011ferinde olmayan eleman bulunamaz. O halde bu k\u00fcmeler birbirine e\u015fitmi\u015f, yani sadece bir tane bo\u015fk\u00fcme varm\u0131\u015f.<\/p>\n – G\u00fczel\u2026 Ama kan\u0131tta k\u00fcmelerin e\u015fitli\u011finden s\u00f6z ettin, bunu bir aksiyom olarak belirtseydin iyi olurdu. \u015eimdi size olduk\u00e7a garip gelebilecek bir teoremden s\u00f6z edeyim: Bo\u015fk\u00fcmenin her eleman\u0131\u00a0<\/em>\u03c0\u2019ye e\u015fittir.<\/em><\/p>\n – Ama nas\u0131l olur, bo\u015fk\u00fcmenin hi\u00e7 eleman\u0131 yok ki!<\/p>\n – \u0130\u015fte tam da bu y\u00fczden bu \u00f6nerme bir teorem, yani kan\u0131tlanabilir.<\/p>\n – Anlamad\u0131m, hi\u00e7 eleman\u0131 olmayan bo\u015fk\u00fcmenin her eleman\u0131 \u03c0\u2019<\/em>ye nas\u0131l e\u015fit olabilir? Biraz de\u011fil \u00e7ok tuhaf!<\/p>\n – Evet, tuhaf g\u00f6r\u00fcn\u00fcyor ama kan\u0131t\u0131 var. \u015e\u00f6yle ki: Yine \u00e7eli\u015fki y\u00f6ntemini kullanal\u0131m. Varsayal\u0131m ki teorem yanl\u0131\u015f, yani bo\u015fk\u00fcmenin her eleman\u0131 \u03c0\u2019<\/em>ye e\u015fit de\u011fil. Bu durumda bo\u015fk\u00fcmede \u03c0\u2019<\/em>ye e\u015fit olamayan en az bir eleman vard\u0131r. Ama bo\u015fk\u00fcmede hi\u00e7 eleman yok! Bu y\u00fczden bo\u015fk\u00fcmede \u03c0\u2019<\/em>ye e\u015fit olmayan bir eleman olamaz. B\u00f6ylece ba\u015flang\u0131\u00e7ta varsayd\u0131\u011f\u0131m\u0131z \u00f6nermenin olumsuzu olan bir sonuca ula\u015ft\u0131k. O halde bo\u015fk\u00fcmenin her eleman\u0131 \u03c0\u2019<\/em>ye e\u015fit!<\/p>\n – Hocam, tamam ama \u201cmatematik bizi kand\u0131r\u0131yor\u201d. Hem bo\u015fk\u00fcmenin hi\u00e7 eleman\u0131 yok diyor, sonra da kalk\u0131p her eleman\u0131n \u03c0\u2019<\/em>ye e\u015fit oldu\u011funu s\u00f6yl\u00fcyor. O zaman bo\u015fk\u00fcmenin her eleman\u0131n herhangi bir say\u0131ya e\u015fit oldu\u011fu da kan\u0131tlanm\u0131\u015f oluyor.<\/p>\n – Elbette, ama bo\u015fk\u00fcmenin hi\u00e7 eleman\u0131 yok! Bu y\u00fczden her eleman\u0131 bir\u00e7ok \u00f6zelli\u011fi sa\u011flar. G\u00fcnl\u00fck mant\u0131k \u00fczerinden bakt\u0131\u011f\u0131m\u0131zda akl\u0131m\u0131za yatmayan b\u00f6ylesi \u00f6nermeler matemati\u011fin tutarl\u0131l\u0131\u011f\u0131 ve ileriki ad\u0131mlar i\u00e7in gereklidir.<\/p>\n – Bu duruma g\u00fcnl\u00fck hayattan \u015f\u00f6yle bir \u00f6rnek geliyor akl\u0131ma: \u00d6rne\u011fin ben Ekim ay\u0131nda yap\u0131lacak s\u00f6zl\u00fc s\u0131navlardan en az 50 alaca\u011f\u0131m iddias\u0131nda bulunuyorum, ama Ekim\u2019de s\u00f6zl\u00fc s\u0131nav yap\u0131lm\u0131yor. Bu durumda iddiam ge\u00e7erlili\u011fini kaybetmiyor, yani bo\u015fk\u00fcmenin her eleman\u0131 herhangi bir reel say\u0131ya e\u015fit!<\/p>\n – Olabilir, ama matematik d\u00fcnyas\u0131 soyuttur. Bu y\u00fczden ger\u00e7ekler d\u00fcnyas\u0131ndaki olaylarla bir teoremi a\u00e7\u0131klamak do\u011fru de\u011fildir.\u00a0 Bu teoremin \u00f6nemini, k\u00fcmeler kuram\u0131n\u0131, say\u0131lar\u0131n in\u015fas\u0131n\u0131 inceledi\u011fimizde daha iyi anlayabiliriz. \u00d6rne\u011fin bo\u015f olmayan bir A k\u00fcmesi \u00fczerinde tan\u0131ml\u0131 ba\u011f\u0131nt\u0131lar\u0131n aras\u0131nda bo\u015f k\u00fcme ({}) de vard\u0131r ve bu ba\u011f\u0131nt\u0131 simetri, ters simetri, ge\u00e7i\u015fme gibi \u00f6zellikleri sa\u011flar. Bu \u00f6rnekte de g\u00f6rd\u00fc\u011f\u00fcm\u00fcz gibi matematik d\u00fcnyas\u0131nda dola\u015ft\u0131k\u00e7a bo\u015fk\u00fcmenin her eleman\u0131n bir\u00e7ok \u00f6zelli\u011fi sa\u011flad\u0131\u011f\u0131 \u00f6nermesinin \u00f6nemini daha iyi anlayabiliriz. Nas\u0131l ki, i\u00e7meden karadut \u015ferbetinin tad\u0131n\u0131n ne oldu\u011funu bilemiyoruz, matematik d\u00fcnyas\u0131na girmeden de bir teoremin \u00f6nemini anlayamay\u0131z. Salt sezgilerimiz bizi aldatabilir, matematiksel d\u00fc\u015f\u00fcnmenin yolundan ayr\u0131labiliriz.<\/p>\n Zek\u00e2 sorular\u0131ndaki matematiksel d\u00fc\u015f\u00fcnme<\/strong><\/p>\n Ge\u00e7en say\u0131da baz\u0131 zek\u00e2 testlerinden se\u00e7ti\u011fim sorular\u0131 payla\u015fm\u0131\u015ft\u0131m. A\u015fa\u011f\u0131da bu sorular\u0131n yan\u0131tlar\u0131n\u0131 veriyoruz.<\/p>\n Bilinmeyen nedir? Verilenler neler? Ko\u015fullar nedir? Bu sorularla ve baz\u0131 s\u0131ra d\u0131\u015f\u0131 ad\u0131mlarla ke\u015ffe \u00e7\u0131kmak zek\u00e2 sorular\u0131yla matemati\u011fin ortak alanlar\u0131 oldu\u011fundan a\u015fa\u011f\u0131da verilen yan\u0131tlar\u0131n \u00e7o\u011funda matematiksel d\u00fc\u015f\u00fcnmenin ipu\u00e7lar\u0131n\u0131 bulmak m\u00fcmk\u00fcn.<\/p>\n Soru 1.<\/strong> \u0130kiz karde\u015f olan A <\/em>ve B<\/em> gibi iki ki\u015fiden A<\/em> do\u011fum g\u00fcn\u00fcn\u00fc kutlad\u0131ktan 2 g\u00fcn sonra B<\/em> do\u011fum g\u00fcn\u00fcn\u00fc kutluyor. Bu nas\u0131l olur?<\/p>\n Yan\u0131t<\/strong>. A<\/em>, 28 \u015eubat gece yar\u0131s\u0131 12\u2019ye 5 kala, B<\/em>\u2019de 12\u2019yi 5 ge\u00e7e do\u011far, dolay\u0131s\u0131yla B<\/em>\u2019nin do\u011fum tarihi 1 Mart\u2019t\u0131r. Soruda s\u00f6z\u00fc edilen do\u011fum g\u00fcnlerinin kutland\u0131\u011f\u0131 y\u0131l \u015eubat 29 \u00e7ekti\u011finden B<\/em>, do\u011fum g\u00fcn\u00fcn\u00fc A<\/em>\u2019dan iki g\u00fcn sonra 1 Mart\u2019ta kutlar.<\/p>\n Soru 2.<\/strong> Ocak \u2013 \u015eubat\u00a0 \u2013 Nisan \u2013 Temmuz \u2013 Kas\u0131m \u2013 Nisan – ?<\/p>\n Yukar\u0131da aylar bir kurala g\u00f6re s\u0131ralanm\u0131\u015ft\u0131r. Soru i\u015fareti yerine hangi ay gelmeli?<\/p>\n Yan\u0131t.<\/strong> Ocak\u2019tan \u015eubat\u2019a ge\u00e7erken ay atlanmam\u0131\u015f, \u015eubat\u2019tan Nisan\u2019a ge\u00e7erken bir ay (Mart) atlanm\u0131\u015f, Nisan\u2019dan Temmuz\u2019a iki ay (May\u0131s, Haziran), Temmuz\u2019dan Kas\u0131m\u2019a \u00fc\u00e7 ay (A\u011fustos, Eyl\u00fcl, Ekim), Kas\u0131m\u2019dan Nisan\u2019a d\u00f6rt ay (Aral\u0131k, Ocak, \u015eubat, Mart) atlanarak bir \u00f6r\u00fcnt\u00fc olu\u015fturulmu\u015f. Nisan\u2019dan sonra be\u015f ay atlan\u0131rsa yan\u0131t Ekim olur.<\/p>\n Soru 3.<\/strong> Bir torbada 50 adet bilye vard\u0131r. Bu bilyelerden en az birinin mavi renkte ve ayr\u0131ca, rastgele \u00e7ekilecek iki bilyeden en az birinin kesinlikle ye\u015fil renkte oldu\u011funu biliyoruz. Torbada ka\u00e7 ye\u015fil bilye vard\u0131r?<\/p>\n Yan\u0131t.\u00a0<\/strong>49 ye\u015fil bilye vard\u0131r.<\/p>\n Soru 4.<\/strong> Bir torbada 4 k\u0131rm\u0131z\u0131, 5 mavi, 6 ye\u015fil bilye vard\u0131r. Bu torbadan en az ka\u00e7 bilye \u00e7ekilirse kesinlikle 2 k\u0131rm\u0131z\u0131 veya 3 ye\u015fil bilye \u00e7ekilmi\u015f olur?<\/p>\n Yan\u0131t.\u00a0<\/strong>2 k\u0131rm\u0131z\u0131 veya 3 ye\u015fil bilye \u00e7ekmeyi garantilemek istedi\u011fimizden ilk \u00e7ekti\u011fimiz 5 bilyenin mavi oldu\u011funu varsaymal\u0131y\u0131z. Sonras\u0131nda 2 bilye \u00e7ekersek, 2\u2019si de k\u0131rm\u0131z\u0131 gelebilir ama kesin de\u011fil, 2\u2019si de ye\u015fil olabilir. Bu durumda istenen sa\u011flanmaz. 5 mavinin ard\u0131ndan 3 bilye \u00e7ekersek de istenen olmayabilir ( 1 k\u0131rm\u0131z\u0131 2 ye\u015fil). Ama 4 bilye \u00e7ekti\u011fimizde 2 k\u0131rm\u0131z\u0131 veya 3 ye\u015fil garantilenir. Bu y\u00fczden en az 5 + 4 = 9 bilye \u00e7ekilmeli.<\/p>\n Soru 5.<\/strong> 4, 5, 4, 5, 5, 7, 6, 7, 5, 4, 5, ? dizisinde soru i\u015fareti yerine hangi say\u0131 gelmelidir?<\/p>\n Yan\u0131t.<\/strong>Bu dizideki say\u0131lar Ocak, \u015eubat, Mart, Nisan, May\u0131s, Haziran, Temmuz A\u011fustos, Eyl\u00fcl, Ekim, Kas\u0131m s\u00f6zc\u00fcklerindeki harf say\u0131s\u0131n\u0131 g\u00f6steriyor. Aral\u0131k s\u00f6zc\u00fc\u011f\u00fcndeki harf say\u0131s\u0131 sorunun yan\u0131t\u0131d\u0131r: 6.<\/p>\n Soru 6.<\/strong> 30a<\/em> + 28b<\/em> + 31c<\/em> = 365 e\u015fitli\u011fini sa\u011flayan a<\/em>, b<\/em>, c<\/em> pozitif tamsay\u0131lar\u0131n\u0131 bulunuz.<\/p>\n Yan\u0131t.<\/strong> 4 ay 30, 1 ay 28, 7 ay 31 \u00e7ekti\u011finden a = 7, b = 1, c = 7.<\/p>\n Soru 7.<\/strong> Bir k\u0131z\u0131n erkek karde\u015flerinin say\u0131s\u0131 k\u0131z karde\u015flerinin say\u0131s\u0131na e\u015fittir. \u00d6te yandan her erke\u011fin k\u0131z karde\u015flerinin say\u0131s\u0131 erkek karde\u015flerinin say\u0131s\u0131n\u0131n iki kat\u0131d\u0131r. Bu ailede ka\u00e7 k\u0131z ka\u00e7 erkek karde\u015f vard\u0131r?<\/p>\n Yan\u0131t.<\/strong>Sorunun ilk c\u00fcmlesinden bu ailede k\u0131z karde\u015flerin say\u0131s\u0131n\u0131n erkek karde\u015f say\u0131s\u0131ndan 1 fazla oldu\u011funu anl\u0131yoruz. \u0130kinci c\u00fcmle, k\u0131z karde\u015f say\u0131s\u0131n\u0131n erkek karde\u015f say\u0131s\u0131n\u0131n 1 eksi\u011finin 2 kat\u0131na e\u015fit oldu\u011funu s\u00f6yl\u00fcyor. Buradan 4 k\u0131z, 3 erkek karde\u015f oldu\u011fu sonucuna ula\u015f\u0131r\u0131z.<\/p>\n Soru 8.<\/strong>A<\/em>, B<\/em>, C<\/em>, D<\/em> ve E<\/em> gibi 5 \u00f6\u011frenci m\u00fcd\u00fcr odas\u0131n\u0131n cam\u0131n\u0131 k\u0131rd\u0131klar\u0131 i\u00e7in m\u00fcd\u00fcr taraf\u0131ndan sorgulanmaktad\u0131rlar. Cam\u0131 k\u0131ran topa kimin vurdu\u011funu \u00f6\u011frenmek isteyen m\u00fcd\u00fcre \u00f6\u011frencilerden sadece 3\u2019\u00fc do\u011fruyu s\u00f6yl\u00fcyor. A\u015fa\u011f\u0131daki ifadelere g\u00f6re do\u011fruyu s\u00f6yleyen 3 \u00f6\u011frenci hangileridir?<\/p>\n A<\/em>: \u201cTopa D<\/em> vurdu.\u201d<\/p>\n B<\/em>: \u201cBen vurmad\u0131m.\u201d<\/p>\n C<\/em>: \u201cTopa E <\/em>vurmad\u0131.\u201d<\/p>\n D<\/em>: \u201cA<\/em> yalan s\u00f6yl\u00fcyor.\u201d<\/p>\n E<\/em>: \u201cB<\/em> do\u011fru s\u00f6yl\u00fcyor.\u201d<\/p>\n Yan\u0131t.<\/strong>A\u015fa\u011f\u0131daki tabloda s\u00fctunlar \u00f6\u011frencilerin ifadelerine g\u00f6re olas\u0131 do\u011fru olabilecek durumlar\u0131n + i\u015faretiyle g\u00f6sterilmesiyle olu\u015fturulmu\u015ftur. Tablodan topa vuran \u00f6\u011frencinin E<\/em> oldu\u011funu anlayabiliriz, \u00e7\u00fcnk\u00fc soruda sadece 3 \u00f6\u011frencinin do\u011fruyu s\u00f6yledi\u011fi veriliyor ve tabloda sadece E<\/em>\u2019nin bulundu\u011fu sat\u0131rda 3 tane + i\u015fareti var. Demek ki 4 + i\u015fareti olan s\u00fctunlardaki \u00f6\u011frenciler do\u011fruyu s\u00f6ylemi\u015f: B<\/em>, D<\/em> ve E<\/em>.<\/p>\n Sade kahveyle dolu bir bardak kahvenin \u00fc\u00e7te birini i\u00e7iyorsunuz.<\/p>\n Sonras\u0131nda, barda\u011fa i\u00e7ti\u011finiz kahve kadar (bardaktaki sade kahvenin \u00fc\u00e7te biri kadar) s\u00fct koyup kar\u0131\u015ft\u0131r\u0131yorsunuz.<\/p>\n Daha sonra bardaktaki s\u00fctl\u00fc kahve kar\u0131\u015f\u0131m\u0131n\u0131n yar\u0131s\u0131n\u0131 i\u00e7iyorsunuz.<\/p>\n Son olarak i\u00e7mi\u015f oldu\u011funuz s\u00fctl\u00fc kahve kar\u015f\u0131m\u0131 kadar miktarda s\u00fct\u00fc barda\u011fa bo\u015falt\u0131yor ve kar\u0131\u015ft\u0131r\u0131yorsunuz.<\/p>\n Bu kez, elde etti\u011finiz s\u00fctl\u00fc kahve kar\u0131\u015f\u0131m\u0131n\u0131n alt\u0131 da birini i\u00e7iyorsunuz.<\/p>\n Daha sonra az \u00f6nce i\u00e7mi\u015f oldu\u011funuz miktar kadar s\u00fct\u00fc barda\u011fa bo\u015falt\u0131yor ve kar\u0131\u015ft\u0131r\u0131yorsunuz. Ve nihayet bardaktaki kar\u0131\u015f\u0131m\u0131n t\u00fcm\u00fcn\u00fc i\u00e7iyorsunuz.<\/p>\n Soru \u015f\u00f6yle: Sonu\u00e7ta toplam olarak daha \u00e7ok s\u00fct m\u00fc, daha \u00e7ok kahve mi i\u00e7tiniz?<\/p>\n Yan\u0131t.\u00a0<\/strong>S\u00fct ve kahveyi ayn\u0131 miktarda i\u00e7mi\u015f olursunuz. \u015e\u00f6yle ki: Ba\u015flang\u0131\u00e7ta bardakta sadece kahve var. Sonras\u0131ndaki ad\u0131mlarda barda\u011fa d\u00f6kt\u00fc\u011f\u00fcn\u00fcz s\u00fct miktar\u0131na bakal\u0131m; s\u0131ras\u0131yla dolu barda\u011f\u0131n 1\/3\u2019u, 1\/2\u2019si ve 1\/6\u2019s\u0131. Bu say\u0131lar\u0131n toplam\u0131 1. Demek ki barda\u011fa bir bardak dolusu s\u00fct d\u00f6km\u00fc\u015fs\u00fcn\u00fcz. Ba\u015flang\u0131\u00e7ta bardak kahveyle dolu oldu\u011funa g\u00f6re sonu\u00e7ta bir bardak kahve, bir bardak s\u00fct i\u00e7mi\u015f olursunuz.<\/p>\n Soru 10.<\/strong> Her g\u00fcn saat 17.00\u2019da trenden inen bir ki\u015fiyi istasyona tam 17.00\u2019da gelen e\u015fi kar\u015f\u0131l\u0131yor ve birlikte otomobille eve d\u00f6n\u00fcyorlar. Bir g\u00fcn bu ki\u015fi trenden saat 16.00\u2019da iniyor, eve do\u011fru y\u00fcr\u00fcmeye ba\u015fl\u0131yor; yolda, istasyona erken geldi\u011finden haberi olmayan ve otomobille 60 km\/s h\u0131zla istasyona do\u011fru gelen e\u015fine rastl\u0131yor, otomobille eve d\u00f6n\u00fcyorlar.\u00a0 Bu ki\u015fi eve her zamankinden 10 dakika \u00f6nce vard\u0131\u011f\u0131na g\u00f6re y\u00fcr\u00fcrken ki h\u0131z\u0131 ka\u00e7 km\/s tir?<\/p>\n Yan\u0131t.<\/strong> E\u015fini kar\u015f\u0131lamak i\u00e7in otomobille gelen ki\u015finin her g\u00fcn saat 17.00\u2019da istasyonda olmas\u0131 sorunun p\u00fcf noktas\u0131. Eve 10 dakika erken d\u00f6nd\u00fcklerine g\u00f6re otomobil kar\u015f\u0131la\u015ft\u0131klar\u0131 yerden istasyona 10 dakikada gidip gelebiliyor. Demek ki otomobil kar\u015f\u0131la\u015ft\u0131klar\u0131 yerden istasyona 5 dakikada gidebiliyor, yani 16.55\u2019te kar\u015f\u0131la\u015fm\u0131\u015flar ve trenden inen ki\u015fi 55 dakika y\u00fcr\u00fcm\u00fc\u015f. Otomobilin 5 dakikada gitti\u011fi yolu y\u00fcr\u00fcyen ki\u015fi 55 dakikada gitmi\u015f. O halde y\u00fcr\u00fcyen ki\u015finin h\u0131z\u0131 60\/11 km\/s\u2019tir.<\/p>\n Soru 11.<\/strong> Her biri di\u011ferlerine yollarla ba\u011fl\u0131 olan A, B, C gibi 3 \u015fehirden A\u2019dan B\u2019ye (C\u2019den ge\u00e7en yollar da dahil) 82 farkl\u0131 \u015fekilde ula\u015f\u0131labiliyor. B\u2019den C\u2019ye ( A\u2019dan ge\u00e7enler de dahil) 62 farkl\u0131 yol var. A\u2019dan C\u2019ye (B\u2019den ge\u00e7enler de dahil) en az ka\u00e7 farkl\u0131 yoldan gidilebilir?<\/p>\n Yan\u0131t.<\/strong> A\u2019dan B\u2019ye, B\u2019den C\u2019ye ve C\u2019den A\u2019ya do\u011frudan gidilebilecek yollar\u0131n say\u0131s\u0131n\u0131 s\u0131ras\u0131yla a, b ve c ile g\u00f6sterelim. A\u2019dan B\u2019ye do\u011frudan ya da C\u2019ye u\u011frayarak gidilebilir. Bu durumda A\u2019dan B\u2019ye giden toplam yol say\u0131s\u0131 a + bc = 82 olur. Benzer \u015fekilde B\u2019den C\u2019ye giden toplam yol say\u0131s\u0131 b + ac = 62\u2019dir. Bu e\u015fitlikleri taraf tarafa \u00e7\u0131kar\u0131rsak<\/p>\n (b\u2013a)(c\u20131) = 20<\/p>\n e\u015fitli\u011fi elde edilir. Bu e\u015fitlikte c\u20131, 20\u2019nin pozitif tam b\u00f6lenleridir. Bu b\u00f6lenlere g\u00f6re c\u2019nin alabilece\u011fi de\u011ferler 2, 3, 5, 6, 11, 21\u2019dir, ancak c\u2019ye 5, 6 ve 21 de\u011ferlerini verdi\u011fimizde a ve b tamsay\u0131 olmuyor. Geriye kalan 2, 3 ve 11 say\u0131lar\u0131 aras\u0131nda A\u2019dan C\u2019ye gidilebilecek yol say\u0131s\u0131n\u0131 en az yapan c de\u011feri 11\u2019dir. Buradan b = 7, a = 5 olur. A\u2019dan C\u2019ye gidilebilecek en az yol say\u0131s\u0131 ise 11 + 7 \u00d7 5 = 46 bulunur.<\/p>\n Soru 12.<\/strong> \u00d6\u011fretmen \u00f6\u011frencilerine \u015f\u00f6yle bir duyuru yapar: \u201c\u00d6n\u00fcm\u00fczdeki hafta s\u0131nav olacaks\u0131n\u0131z ama hangi g\u00fcn s\u0131nav olaca\u011f\u0131n\u0131z\u0131 bir \u00f6nceki g\u00fcn bilemeyeceksiniz\u201d. \u00d6\u011fretmen hangi g\u00fcn s\u0131nav yapm\u0131\u015f olabilir?<\/p>\n Yan\u0131t.<\/strong>\u00d6\u011fretmen hi\u00e7bir g\u00fcn s\u0131nav yapamaz; \u00e7\u00fcnk\u00fc e\u011fer Per\u015fembe g\u00fcn\u00fc s\u0131nav yap\u0131lmam\u0131\u015fsa \u00f6\u011frenciler Cuma g\u00fcn\u00fc s\u0131nav yap\u0131laca\u011f\u0131n\u0131 anlarlar. Bu y\u00fczden haftan\u0131n son g\u00fcn\u00fcnde (Cuma) s\u0131nav yap\u0131lamaz. Bu ak\u0131l y\u00fcr\u00fctmeyle geriye do\u011fru gidersek, Per\u015fembe, \u00c7ar\u015famba ve di\u011fer g\u00fcnler i\u00e7in de s\u0131nav yap\u0131lamayaca\u011f\u0131 sonucuna ula\u015f\u0131r\u0131z. Bu bir paradokstur.<\/p>\n Soru 13.<\/strong><\/p>\n Yukar\u0131daki \u015fekilde 12 kibrit \u00e7\u00f6p\u00fcyle 6 e\u015fkenar \u00fc\u00e7genden olu\u015fan bir alt\u0131gen \u00e7izilmi\u015ftir.<\/p>\n A) 4 kibrit \u00e7\u00f6p\u00fcn\u00fc oynatarak 3 e\u015fkenar \u00fc\u00e7gen olu\u015fturunuz.<\/p>\n B) 3 kibrit \u00e7\u00f6p\u00fcn\u00fc oynatarak 4 e\u015f paralelkenar olu\u015fturunuz.<\/p>\n C) 4 kibrit \u00e7\u00f6p\u00fcn\u00fc oynatarak 4 e\u015f paralelkenar olu\u015fturunuz.<\/p>\n Yan\u0131t.<\/strong><\/p>\n <\/p>\n Soru 14. <\/strong>6 kibrit \u00e7\u00f6p\u00fcyle 4 e\u015fkenar \u00fc\u00e7gen nas\u0131l olu\u015fturulur?<\/p>\n Yan\u0131t. <\/strong>E\u015fkenar \u00fc\u00e7gen piramit.<\/p>\n Soru 15.<\/strong><\/p>\n Yukar\u0131daki \u015fekilde toplam \u00fc\u00e7gen say\u0131s\u0131 ka\u00e7t\u0131r? Bu hesaplamay\u0131 yapabilmek i\u00e7in bir y\u00f6ntem geli\u015ftiriniz. ( Kombinasyon form\u00fcl\u00fcn\u00fc kullanmaman\u0131z gerekiyor.)<\/p>\n Yan\u0131t. <\/strong>\u015eekildeki \u00fc\u00e7gen say\u0131s\u0131n\u0131 hesaplamak i\u00e7in bir\u00e7ok y\u00f6ntem bulunabilir. Bu yollardan birisi \u015f\u00f6yle: \u00dc\u00e7genin yataydaki kenar\u0131ndan (taban\u0131) ba\u015flayarak yukar\u0131ya do\u011fru olu\u015fan \u00fc\u00e7genleri sayabiliriz. En a\u015fa\u011f\u0131da kalan ve tepe noktas\u0131 tabana en yak\u0131n olan 1 \u00fc\u00e7gen vard\u0131r. Sonra, tepe noktas\u0131 en a\u015fa\u011f\u0131daki \u00fc\u00e7genin hemen \u00fcst\u00fcnde olan \u00fc\u00e7gendeki toplam \u00fc\u00e7gen say\u0131s\u0131 bulunur:<\/p>\n 1\u00d71+2\u00d72+3\u00d71 = 8.<\/p>\n Ayn\u0131 yolla tepe noktas\u0131 a\u015fa\u011f\u0131dan yukar\u0131ya do\u011fru \u00fc\u00e7\u00fcnc\u00fc s\u0131rada olan \u00fc\u00e7genin i\u00e7indeki \u00fc\u00e7genlerin say\u0131s\u0131 bulunur:<\/p>\n 1\u00d71+2\u00d72+3\u00d73+4\u00d72+5\u00d71 = 27.<\/p>\n Bu yolla \u015fekilde verilen \u00fc\u00e7genin tepe noktas\u0131 dikkate al\u0131n\u0131rsa arad\u0131\u011f\u0131m\u0131z say\u0131: Dikkat edilirse \u00fc\u00e7gen say\u0131lar\u0131 \u00a0ve \u2019\u00fcn k\u00fcp\u00fc oluyor. Dolay\u0131s\u0131yla genel bir \u00e7al\u0131\u015fma yap\u0131larak toplam \u00fc\u00e7gen say\u0131s\u0131n\u0131 veren form\u00fcl, \u00fc\u00e7genin yan kenarlar\u0131n\u0131n b\u00f6l\u00fcnt\u00fc say\u0131s\u0131n\u0131n k\u00fcp\u00fc olabilir.<\/p>\n","protected":false},"excerpt":{"rendered":" \u0130ki \u00f6\u011frenci at\u0131\u015f\u0131yorlard\u0131: – Sen bo\u015fk\u00fcmesin, – Ben de\u011fil, as\u0131l bo\u015fk\u00fcme sensin \u00d6\u011fretmen araya girdi: – \u00dcz\u00fclmeyin, ikiniz de bo\u015fk\u00fcme de\u011filsiniz, ama bo\u015fk\u00fcmeye haks\u0131zl\u0131k ediyorsunuz. Sadece bir tane bo\u015fk\u00fcme var. Oysa bir ve birden \u00e7ok eleman\u0131 olan k\u00fcmelerin s\u00fcr\u00fcs\u00fcne bereket. Bu y\u00fczden bo\u015fk\u00fcme di\u011fer k\u00fcmelere g\u00f6re daha k\u0131ymetli! Bu diyalogun ard\u0131ndan koyu bir bo\u015fk\u00fcme sohbeti […]<\/p>\n","protected":false},"author":375,"featured_media":13954,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[178,25,514],"tags":[208,1008,1572],"class_list":["post-16978","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-141-sayi","category-matematik","category-matematik-sohbetleri","tag-matematik","tag-matematik-sohbetleri","tag-problemler"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/16978","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=16978"}],"version-history":[{"count":0,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/16978\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media\/13954"}],"wp:attachment":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media?parent=16978"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/categories?post=16978"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/tags?post=16978"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/01\/matematik141-tablo.jpg\")
\nSoru 9.<\/strong> Kahve i\u00e7erek d\u00fc\u015f\u00fcnmeyi seven okurlar\u0131n ho\u015flanaca\u011f\u0131 bir soru!<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/01\/matematik141-tablo-3.jpg\")
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\n1\u00d71+2\u00d72+3\u00d73+4\u00d74+5\u00d73+6\u00d72+7\u00d71 = 64.<\/p>\n