Concept of Reminders (Part 4)

Duration: 18 min

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Inside: a video lesson and guided study material.

Module outline

  1. About the Course: About the Course
  2. Section 1 | Language (Hindi): भाषा, लिपि और वर्ण, वर्तनी, वर्ण-विन्यास, अनुस्वार, अनुनासिक, चंद्रबिंदु और ‘र’ के रूप, शब्द, पद और शब्द-भेद, संज्ञा और सर्वनाम, पद-भेद और रूप-विकार, लिंग, वचन, पुरुष और कारक, काल, वाच्य, वाक्य और वाक्य-भेद, संधि, समास, विलोम शब्द, पर्यायवाची शब्द, सामान्य अशुद्धियाँ, अनेक शब्दों के लिए एक शब्द, शब्द-भंडार, अपठित बोध, Multi-Topic Content (Cross-Concept Material), वर्णमाला
  3. Section 1 | Language (उर्दू भाषा — 22 प्रश्न (निश्चित)): رسم الخط، حروفِ تہجی اور املا / Script, Alphabet and Spelling, اسم، ضمیر اور صفت / Nouns, Pronouns and Adjectives, الفاظ، محاورے اور ضرب الامثال / Vocabulary, Idioms and Proverbs, تفہیمِ عبارت، انشا اور ترجمہ / Comprehension, Writing and Translation, Comprehension
  4. Section 1 | Language (বাংলা / बांग्ला भाषा — 22 प्रश्न (निश्चित)): শব্দার্থ, বাগধারা ও ভাষাশুদ্ধি / Meaning, Idioms and Correct Usage, বাক্য ও বাচ্য / Sentences and Voice, শব্দগঠন ও শব্দভান্ডার / Word Formation and Vocabulary, অনুবাদ, বোধপরীক্ষণ ও রচনা / Translation, Comprehension and Composition
  5. Section 1 | Language (English): Vocabulary, Noun, Subject Verb Agreement (Verb Noun Agreement), Adjectives, Verbs and Tenses, Pronoun, Verb, Adverb, Prepositions and Conjunctions, Parts of Speech, Determiner, Conjunction, Modals, Transformation of Sentences, Direct and Indirect Speech, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Construction, Sentence Structure and Common Errors, Reading Comprehension, Cloze Test (Cloze Passage), Verbal Analogy (Word Based Analogy), Facts, Inference and Judgements
  6. Section 2 | Primary Maths & Reasoning: अनुपात और समानुपात / Ratio and Proportion, Divisibility Rules, मिश्रण और एलिगेशन / Mixture and Alligation, Permutation and Combination, संख्या पद्धति / Number System, महत्तम समापवर्तक और लघुत्तम समापवर्त्य / HCF and LCM, Sequence and Series (Series), समय, चाल और दूरी / Time, Speed and Distance, श्रेणी / Series, कूट लेखन एवं कूट वाचन / Coding-Decoding, अशाब्दिक तर्क / Non-Verbal Reasoning, प्रतिशत / Percentage, औसत / Average, लाभ, हानि और बट्टा / Profit, Loss and Discount, साधारण और चक्रवृद्धि ब्याज / Simple and Compound Interest, समय और कार्य / Time and Work, घन क्षेत्रमिति / Solid Mensuration, Age Problems, Arithmetic, सरलीकरण, घातांक और करणी / Simplification, Indices and Surds, Algebra, Probability, Trigonometry, Syllogisms, बैठक व्यवस्था एवं पहेलियाँ / Seating Arrangement and Puzzles, रक्त संबंध / Blood Relations, दिशा एवं दूरी / Direction and Distance, Analytical Reasoning (Counting Figures Reasoning), गणितीय संक्रियाएँ एवं लुप्त संख्या / Mathematical Operations and Missing Numbers, कैलेंडर एवं घड़ी / Calendar and Clock, Clock, घन, पासा एवं आकृतियों की गिनती / Cubes, Dice and Figure Counting, वेन आरेख, न्याय वाक्य एवं असमानता / Venn Diagrams, Syllogism and Inequalities, Data Arrangement, Inequality, सादृश्यता एवं वर्गीकरण / Analogy and Classification, क्रम एवं रैंकिंग / Order and Ranking, समुच्चय, संबंध, फलन और विविक्त गणित / Sets, Relations, Functions and Discrete Mathematics, संख्या पद्धति और संख्या सिद्धांत / Number Systems and Number Theory, निर्देशांक ज्यामिति, सदिश और त्रिविमीय ज्यामिति / Coordinate Geometry, Vectors and 3D Geometry, बहुपद, समीकरण और सम्मिश्र संख्याएँ / Polynomials, Equations and Complex Numbers
  7. Section 2 | General Awareness (GK + Current Affairs): Polity, वैज्ञानिक उपलब्धियाँ और अनुसंधान / Scientific Achievements and Research, Biology, राज्य विशेष सामान्य ज्ञान / State-Specific General Knowledge, Chemistry, History, Geography, Indian Economy, Environment, समसामयिकी / Current Affairs, Static General Knowledge
  8. Section 3 | Computer System and Org. (CSO): Basics of COA, Floating Point Rep, Cache Memory Organization, Instr Formats & Modes, Programming the Basic Computer, Input Output Organisation, Register Transfer and Microoperations, Control Unit Design
  9. Section 3 | Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, SQL, File Organization & Indexing, Relational Algebra, Relational Calculus, Transaction Management, Concurrency Control
  10. Section 3 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Process Synchronization
  11. Section 3 | Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics
  12. Section 3 | Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing, Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Minimum Spanning Trees, Shortest Path Algos
  13. Section 3 | Theory Of Computation & Compiler Design: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines, Intro to Compilers, Lexical Analysis, Grammar & CFG, Syntax Analysis: Top-Down, Syntax Analysis: Bottom-Up, Semantic Analysis & SDT, Intermediate Code Gen, Code Optimization
  14. Section 3 | Emerging Trends (AI, Societal Ethics & MS Office): कंप्यूटर के मूल सिद्धांत / Computer Fundamentals, Emerging Tech Trends, Society, Law & Ethics, MS Office Applications, व्यापार, विपणन एवं उपभोक्ता संरक्षण / Trade, Marketing and Consumer Protection, साइबर सुरक्षा / Cyber Safety and Security, कंप्यूटर सॉफ़्टवेयर / Computer Software, ऑपरेटिंग सिस्टम / Operating System, कंप्यूटर नेटवर्क / Computer Networks, वर्ड प्रोसेसिंग (MS Word) / Word Processing (MS Word), Terms & Abbreviations, प्रेज़ेंटेशन (MS PowerPoint) / Presentation Software (MS PowerPoint)
  15. Section 3 | Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Maintenance and Configuration Management, Software Testing
  16. Section 3 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  17. Section 3 | OOPS, C++, JAVA, HTML: OOPS with C++, Java, HTML Basics, CSS
  18. Section 3 | Java: Java Fundamentals, Array, String, IO & References, OOPS with Java
  19. Section 3 | Python: Introduction to Python, Tokens, Variables, Data Types, I/O, Operators & Expressions, Control Flow & Looping, Collections & Strings, Functions & Modules, File Handling & Data Analysis
  20. Section 3 | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation
  21. Section 3 | Multimedia: Multimedia
  22. BPSC Previous Year Papers:
  23. BPSC Full Mock Tests:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video, presented by Yash Jain from Knowledge Gate, is a comprehensive tutorial on solving remainder problems in number systems, a common topic in competitive exams. The video begins with an introduction to the concept of remainders, using a simple division example (7 ÷ 2 = 3 R1) to establish the foundational idea. The main content consists of two complex problems. The first problem involves the sum of the reciprocals of the first 13 natural numbers, expressed as a fraction x/13!. The instructor demonstrates that to find the remainder when x is divided by 11, one must analyze the numerator of the sum after a common denominator is applied. The key insight is that for any term where the denominator is not divisible by 11, the corresponding numerator term will be divisible by 11, leaving only the term where the denominator is 11 as the only non-zero contributor to the remainder. The second problem asks for the remainder when N = (1! + 2! + 3! + ... + 1000!)^40 is divided by 10. The instructor explains that for n ≥ 5, n! is divisible by 10, so all terms from 5! onwards are 0 modulo 10. This simplifies the sum to (1! + 2! + 3! + 4!)^40 = (1 + 2 + 6 + 24)^40 = 33^40. The final step is to find the remainder of 33^40 when divided by 10, which is equivalent to finding the last digit of 33^40. The video concludes with a summary of the problem-solving techniques, emphasizing the use of modular arithmetic and the simplification of large expressions by identifying terms that become zero under the given modulus.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title card for 'NUMBER SYSTEM' and then transitions to a presentation slide. The slide features the title 'NUMBER SYSTEM' and the tagline 'The mysterious world of numbers...'. The instructor, Yash Jain, is visible in a small window. He introduces the topic of the 'Concept of Remainders' and provides a simple example on the board: '7 ÷ 2 = 3 R1', with an arrow pointing to the '1' and labeling it 'Remainder'. This establishes the basic definition of a remainder as the leftover part of a division.

  2. 2:00 – 5:00 02:00-05:00

    The video presents the first problem: 'The expression sigma(1/n) from n=1 to 13 can also be written as x/13!. What would be the remainder if x is divided by 11?'. The instructor begins to solve this by writing the sum as 1/1 + 1/2 + 1/3 + ... + 1/13. He then multiplies the entire sum by 13! to get x, which results in the expression x = 13!/1 + 13!/2 + 13!/3 + ... + 13!/13. He explains that to find x mod 11, he needs to evaluate each term in the sum modulo 11.

  3. 5:00 – 10:00 05:00-10:00

    The instructor analyzes the terms of the sum x = 13!/1 + 13!/2 + ... + 13!/13 modulo 11. He explains that 13! is divisible by 11, so 13! ≡ 0 (mod 11). For any term where the denominator is not 11, the numerator 13! is divisible by 11, making the entire term 0 modulo 11. The only term that is not divisible by 11 is the one where the denominator is 11, which is 13!/11. He simplifies this term to 13 × 12 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 13 × 12 × 10!. He then states that the remainder is the value of 13 × 12 × 10! modulo 11, which he will calculate in the next step.

  4. 10:00 – 15:00 10:00-15:00

    The instructor calculates the remainder of 13 × 12 × 10! when divided by 11. He uses the property that (p-1)! ≡ -1 (mod p) for a prime p (Wilson's Theorem). Since 11 is prime, 10! ≡ -1 (mod 11). He then simplifies 13 × 12 modulo 11: 13 ≡ 2 (mod 11) and 12 ≡ 1 (mod 11). Therefore, the expression becomes 2 × 1 × (-1) = -2. Since the remainder must be positive, he adds 11 to get 9. The final answer for the first problem is 9. He then transitions to the second problem.

  5. 15:00 – 18:29 15:00-18:29

    The second problem is introduced: 'What is the remainder when N = (1! + 2! + 3! + ... + 1000!)^40 is divided by 10?'. The instructor explains that for n ≥ 5, n! is divisible by 10 (since it contains both 2 and 5 as factors), so n! ≡ 0 (mod 10). This means all terms from 5! to 1000! are 0 modulo 10. The sum simplifies to (1! + 2! + 3! + 4!)^40 = (1 + 2 + 6 + 24)^40 = 33^40. The problem now is to find 33^40 mod 10, which is the last digit of 33^40. He notes that the last digit of 33 is 3, so he needs to find the last digit of 3^40. He then calculates the pattern of the last digit of powers of 3: 3^1=3, 3^2=9, 3^3=7, 3^4=1, and the cycle repeats every 4. Since 40 is divisible by 4, the last digit is 1. The final answer is 1. The video ends with a 'Thank You for Watching' screen.

The video provides a clear and structured lesson on applying modular arithmetic to solve complex remainder problems. It begins with a fundamental concept and progresses to two distinct, challenging problems. The first problem demonstrates the power of simplification by identifying that only one term in a large sum contributes to the remainder when the modulus is a prime number, leveraging Wilson's Theorem. The second problem illustrates the technique of reducing a sum by recognizing that terms beyond a certain point become zero modulo the divisor. The core teaching method is the step-by-step breakdown of complex expressions into manageable parts, using properties of factorials and modular arithmetic to arrive at the solution. The progression from a simple example to two advanced problems effectively builds the viewer's problem-solving skills for competitive exams.

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