TTT

Masalan. n=11, x ixtiyoriy bo'lsin: K1=11, K2=x, K3=1; 1-qadam: 1) K3=K3×K2=1 × x=x va K1=10; 2) K2=K2×K2=x × x=x^2; 3) K1=K1÷2=10÷2=5; 2-qadam: 1) K3=K3×K2=x × x^2=x^3 va K1=4; 2) K2=K2×K2=x^2 × x^2=x^4; 3) K1=K1÷2=4÷2=2; 3-qadam: 1) - - - 2) K2=K2×K2=x^4 × x^4=x^8; 3) K1=K1÷2=2÷2=1; 4-qadam: 1) K3=K3×K2=x^3 × x^8=x^11 va K1=0; 2) K2=K2×K2=x^8 × x^8=x^16; 3) K1=K1÷2=0÷2=0; Izoh: Foydasini n soni kattalashsa sezish mumkin. Masalan 1024 < n < 2048 uchun ketma ket ko'paytirish orqali topsangiz kamida 1025 marta amal bajarish kerak. Yuqoridagi kabi bajarsangiz, har bitta qadamda ko'pi bilan 4 ta amal bajardik demak amallar soni =< 4×([log(2047)]+1) = 44 dan oshmaydi, albatta qo'shimcha hotira hisobiga.

Yechim 8. 1- kalkulyatorda yozilgan qiymatni K1 deb belgilaylik, (shunga o'xshash qolganlarini K2, K3 kabi belgilaymiz) keyinchalik agar K1=K-1 deb yozilsa demak birinchi kalkulyatordagi sondan 1 ni ayirgan bo'lamiz, huddi shunday K1=K1÷2 yozilsa qiymatni 2 ga bo'lgan bo'lamiz. Quyidagi 3 ta amallar guruhini "qadam" deb nomlaylik: 1) agar K1 toq bo'lsa: K3=K3×K2 va K1=K1-1 (demak K1 juft bo'lsa bu bosqich bajarilmaydi) 2) K2=K2×K2; 3) K1=K1÷2; Endi har bir kalkulyatorga quyidagi qiymatlarini yozib olamiz: K1 = n; K2 = x; K3 = 1; va K1>0 bo'lsa yuqoridagi "qadam"ni takrorlayveramiz. Bundan yakuniy natija K3 da hosil bo'ladi. Jami "qadam"lar soni esa taxminan [log(n)]+1 ga teng bo'ladi (bu yerda log - ikki asosga ko'ra logarifm). Izoh: Asosiy g'oya ikkilik sanoq sistemasi: Istalgan natural sonni ikkilik sanoq sistemasida yozishimiz mumkin, bu esa o'sha sonni 2 ning qaysidir darajalari yig'indisi yordamida ifodalash mumkin degani. Masalan 7 = "111" = 2^2 + 2^1 + 2^0. Demak biz x^7 uchun x, x^2 va x^4 ni bilishimiz yetarli ekan, bular esa o'zidan bitta oldingisini o'ziga ko'paytirilishi orqali hosil bo'ladi. Biz K1 dan n ni ikkilik sanoq sistemasiga yoyishda, K2 dan x ni 1, 2, 4, 8, 16, ..., 2^k darajalarini hosil qilishda va K3 dan natijani saqlashda foydalanayapmiz.

Savol: Tasavvur qiling sizda 3 ta kalkulyator bor va ular 4 ta arifmetik amalni (+, -, ×, ÷) istalgan honali sonlar ustida bajara oladi. Bu kalkulyatorlar yordamida "x" haqiqiy sonini natural "n" darajasini nechta qadamda topish mumkin?

#imc2022 #imathuz ⚡️⚡️ 29-Xalqaro Matematika Musobaqasi (IMC) umumiy natijalari eʼlon qilindi! 📣 Universitet talabalari o‘rtasidagi 29-Xalqaro Matematika Musobaqasi (IMC) bo'lib o'tayotganligi haqida xabar berib o'tgandik. Umumiy natijalar e'lon qilindi. O'zbekistonlik talabalar 1ta oltin, 2ta kumush va 3ta bronza medalni qo'lga kiritishdi. Ular 🥇 Jahongir Turayev (O'zMU) oltin, 🥈Jasurbek Imomov(O'zMU) kumush, 🥈Abdushukur Axadov(O'zMU) kumush, 🥉Jaxongir Norboyev (O'zMU) bronza, 🥉Nabixon Nabixonov (O'zMU) bronza, 🥉Rasul Kushnazarov(UrDU) bronza medalni taqdim etdi. Sovrindor talabalarni, ularning ota-onasi va ustozlarini tabriklaymiz!!! ♻️ Olimpiada oid yangiliklarni biz bilan kuzatib boring! Sayt | Telegram | Instagram | Facebook

Ushbu telegram kanaldan istalgan dasturlash tiliga doir kitob topishingiz mumkin.

Zo'r masala ekan.

IMO masalalarini ko'rgach (P1, P2, P4, P5) bu yil ball odatdagidan baland bo'lishini sezganlar ko'p bo'lsa kerak, bu natijalarda aksini topdi.

Ko'paytirish! Kompyuter takomillashgani sari ko'pgina yangi imkoniyatlar paydo bo'ldi, shu sababli nazariyalarni boshidan qurishga qiziqish kuchaygan: Bilamizki, ikkita butun sonni ko'paytirishni odatiy, maktabda o'rgatilgan, usulidan foydalansak "n" xonali sonlarni ko'paytmasini hisoblash uchun "n^2" ga ekvivalent sonda individual amal bajarishimiz kerak. 1960-yilda yangi algoritm orqali bu ishni "n^1.58" qadamda bajarish mumkinligi isbotlangan. 1971-yilda esa yondashuvni o'zgartirish orqali yanayam yaxshi "n×log(n)×log(log(n))" chegaraga kelishgan. Qanchalik foydasi bor deysizmi: hamma biladigan usul orqali 2 milliard xonali ikkita butun sonni ko'paytirishga kompyuter 6 oy vaqt sarflasa, eng so'ngisida 26 sekund sarflaydi! Bu chegarani "n×ln(n)" ga olib kelishga yondashuv bor, lekin yakunlanmagan.

STEP - Bu men biladigan eng zo'r imtihonlardan biri: masalalarni katta qismi tizimli fikrlashni talab qiladi - olimpiada masalasiga o'xshash, ozroq yengilligi bunda yakuniy natijaga erishish uchun qadamlar beriladi. Savollar orasida eski olimpiada masalalari ham bor - shuning uchun tayyorgarlikni endigina boshlagan bo'lsangiz moslashib olish uchun yaxshi. Barcha materiallar uchun ushbu havola dan foydalanishingiz mumkin [STEP past papers].

Ko'proq ma'lumot yoki ro'yxatdan o'tish uchun ushbu havoladan foydalanishingiz mumkin.

IMC Principle Sponsor Huawei Summer Internship Positions for Mathematics Students 2022 Summer Internship Topics in Turing Computation Group 1. Formalisation of modern mathematics. You are assumed to be familiar with pure mathematics, in particular, algebraic geometry, group theory, and topology. We try to figure out the difficulties in formalising and proving theorems in modern mathematics, e.g., the arithmetic of elliptic curves, the cohomological algebra, etc. 2. Learning the skills of calculation. Human creative activities in reasoning and conjecturing, e.g., construction, generalisation, abstraction, analogy, etc., can be considered sort of calculations. We try to formalise these activities and investigate how to train machines to master these skills. You are supposed to be familiar with mathematical logic, functional programming, and the principles of popular machine learning methods. 3. The Derivation Approach to Design Automation. You are supposed to be familiar with functional programming, the principles of programming languages, or computer architecture. The derivation approach is to produce or refine the designs of software or hardware by calculation. We try to produce an extremely simple CPU and some optimised algorithms from specifications through rigorous calculation. The purpose is to figure out whether the refinement process can be formalised and automated. 4. The derivation of arithmetic, algebra, and analysis. You are supposed to be familiar with category theory and functional programming. We try to develop an experimental system to derive the theories of arithmetic, algebra, and analysis from a small set of concepts. 5. Solve math and programming problems. The idea is to create a benchmark of representative theorems, math competition problems, and programming contest problems, based on their requirements for formalising and proving techniques. IMO or IOI background perfectly matches with this topic. 6. Design a meta-calculation environment. We want to figure out the principles and the main components of a meta-calculation environment for mathematicians, including, a user-friendly interface, a proof checker or assistant, and a bunch of automatic mechanisms. The general background of computer science is necessary. The experience of functional programming is preferred. 7. The topological approach to computing and communicating. You are assumed to have very strong geometry and topology background and be very familiar with the computation theory in order to make reasonable contributions on this direction. How to Apply? Please do not hesitate to contact us by email: xiaomeiouyang@huawei.com for any questions or start your application. Huawei Technologies Co., Ltd.

#Hard Savol 7. Tibbiy laboratoriyada 240 ta inyeksiya bor, ulardan faqat bittasi sichqonni uxlata oladi. Agar inyeksiya sichqonga yuborilsa, u roppa-rosa 24 soat ichida uxlaydi (bazilarida 1 soat, lekin aniq 24 soat ichida ta'sir qiladi). Sizda 48 soat vaqt bor va qaysi inyeksiya uxlatish kuchiga ega ekanligini aniqlamoqchisiz. Buning uchun sizga zarur bo'lgan eng kam sondagi lobarotoriya sichqonlari sonini toping.

Yechim 6. Dastlab birinchi ipni ikkala tarafidan va ikkinchi ipni bitta tarafidan yoqamiz. Birinchi ip to'la yonib bo'lgach, bunga aynan 30 daqiqa vaqt ketadi, ikkinchi ipni yoqilmagan uchidan yoqamiz, bu esa 15 daqiqada yonib tugaydi. Demak jami 45 daqiqa bo'ldi.

#Easy Savol 6. Faraz qiling sizda ikkita ip bor, ularning har biri to'la yonib tugashiga aynan 1 soat vaqt ketadi, lekin ip bir xilda yonmaydi: masalan ipning yarmi yonishi uchun 10 daqiqa, qolgan yarmi yonishi uchun esa 50 daqiqa ketishi mumkin. Shu ikki ip yordamida 45 daqiqani o'lchay olasizmi? Sizda istalgan payt olov yoqish imkoniyati bor.

Yechim 5. Toq marta o'zgartirilgan eshiklar ochiq qoladi, ya'ni eshik nomerini bo'luvchilari soni toq bo'lishi kerak. Demak bo'luvchilari soni toq bo'lgan natural sonlarni topishimiz kerak - bunday sonlar faqat to'la kvadrat bo'lganlari (isbotlashga harakat qilib ko'ring). Demak javob 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

#Medium Savol 5. 100 ta eshik bor, dastlab barcha eshiklar yopiq bo'lsin. Biror kishi barcha eshiklardan bir necha marotaba o'tadi va ularni bazilarini quyidagicha o'zgartiradi: agar eshik ochiq bo'lsa yopadi, yopiq bo'lsa ochadi. U birinchi yurishda har bir eshikni o'zgartiradi, ikkinchi yurishda har ikkinchi eshikni o'zgartiradi, ya'ni, 2, 4, 6, 8, ... uchinchi yurishda har uchinchi eshikni o'zgartiradi, ya'ni 3, 6, 9, ... Xuddi shunday, 100-chi yurishda esa 100-eshikni o'zgartiradi. Qaysi eshiklar ochiq qolishini toping.