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şekilde hesaplanıyor."]},"beginTime":74,"endTime":216,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=74&ask_summarization=1"},{"index":2,"title":"Belirli Aralıklarda Sayılar","list":{"type":"unordered","items":["2500 ile 6500 arasında yazılabilir doğal sayı sayısı 870 farklı şekilde hesaplanıyor.","Dört ile bölünebilen doğal sayı yazma olasılığı 370 farklı şekilde hesaplanıyor.","Cevap anahtarı 132 olarak belirtilmiş ancak çözüm sonucunda 100 farklı yol bulunmuş, cevap anahtarı hatalı olabilir."]},"beginTime":216,"endTime":561,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=216&ask_summarization=1"},{"index":3,"title":"Matematik Kitabı Sayfa Numaralandırma Problemi","list":{"type":"unordered","items":["404 sayfadan oluşan bir matematik kitabında kaç adet \"4\" rakamı kullanıldığı hesaplanıyor.","Bir basamaklı sayfalarda \"4\" rakamı bir kere, iki basamaklı sayfalarda \"4\" rakamı onlar basamağına 10, birler basamağına 8 kez kullanılıyor.","Üç basamaklı sayfalarda \"4\" rakamı farklı durumlarda farklı sayıda kullanılıyor ve toplam 86 farklı \"4\" rakamı kullanılmış oluyor."]},"beginTime":567,"endTime":753,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=567&ask_summarization=1"},{"index":4,"title":"Yarışma Jüri Üyeliği Problemi","list":{"type":"unordered","items":["Bir yarışma jüri üyeliği için beş çift arasından bir erkek ve bir kadın seçilecek ve seçilenlerin eş olmayacak.","Beş erkekten birini seçtikten sonra, geriye kalan dört bayandan birini seçmek 5×4=20 farklı ikili 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soruluyor.","a! = 210b! eşitliği sağlanıyor ve 210 asal çarpanlarına ayrılıyor.","a = 7, b = 4; a = 15, b = 13; a = 15, b = 13 olmak üzere 3 farklı durum bulunuyor."]},"beginTime":1036,"endTime":1141,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=1036&ask_summarization=1"},{"index":9,"title":"Faktöriyel ve Üs Problemi","list":{"type":"unordered","items":["6! = 2^n × a olduğuna göre a'nın alabileceği en büyük ve en küçük değerleri toplamı kaçtır soruluyor.","6! = 720 bulunuyor ve 720 = 2^13 × 3 × 5 × 3 olarak asal çarpanlarına ayrılıyor.","a'nın en büyük değeri 720, en küçük değeri 45 olup, toplamları 765 bulunuyor."]},"beginTime":1141,"endTime":1205,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=1141&ask_summarization=1"},{"index":10,"title":"Faktöriyel ve Üs Problemi","list":{"type":"unordered","items":["67! = 15^a × b eşitliğin alabileceği değerlerin toplamı bulunuyor.","67! = 5^13 × 3^13 × 7^13 olarak asal çarpanlarına ayrılıyor.","a'nın alabileceği değerler 1'den 13'e kadar olan sayılar olup, toplamları 220 bulunuyor."]},"beginTime":1205,"endTime":1264,"href":"/video/preview/10138738168541143360?parent-reqid=1769439481601982-471035945191176055-balancer-l7leveler-kubr-yp-vla-226-BAL&text=SS+Science+%26+Math&t=1205&ask_summarization=1"},{"index":11,"title":"Matematik Sorularının Çözümü","list":{"type":"unordered","items":["Bir matematik sorusunda 16 faktöriyel parantezine alınarak işlem yapılmış ve sonucun k türünden eşiti 8k olarak bulunmuştur.","İkinci soruda a/5! + 10a toplamının tanımlı olabilmesi için a'nın 5'e ve 10'a bölünebilmesi gerektiği, bu değerlerin çarpımı 50 olarak hesaplanmıştır.","Videoda bazı zorlayıcı sorular olduğu ve cevap anahtarında bazı hatalar olduğu belirtilmiş, izleyicilerden çözülmesini istediği soru tiplerini 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