Number System (Quick Revision & Practice Questions)

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This video is a comprehensive mathematics lecture on the concepts of multiples and factors, designed for competitive exam preparation. The instructor begins by defining factors and multiples using the number 12 as an example, showing that its factors are {1, 2, 3, 4, 6, 12} and its multiples are {12, 24, 36, 48, ...}. He then introduces the fundamental formula for finding the number of factors of a number N, which is derived from its prime factorization: if N = p^a * q^b * r^c, then the number of factors is (a+1)(b+1)(c+1). The lecture progresses to solve a series of practice problems, including finding the number of factors of a sum, determining the smallest number to multiply to make a perfect square or cube, and solving remainder problems using the formula Dividend = Divisor * Quotient + Remainder. The video concludes with a problem involving the sum of digits of a large number and a unique problem about the square root of a repeating digit sequence.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The video opens with a title card for 'NUMBER SYSTEM' followed by a presentation slide titled 'MULTIPLES AND FACTORS'. The instructor, Yash Jain Sir, begins the lesson by defining factors of a number N as all numbers that divide N completely. He introduces the concept of the number of factors and the product of factors, setting the stage for the main topic.

2. 2:00 – 5:00 02:00-05:00

   The instructor writes the title 'Concept of Multiples & Factors' on the whiteboard. He uses the number 12 as an example, writing '12 = {1, 2, 3, 4, 6, 12}' to list its factors. He then explains that the number of factors of 12 is 6 and that the product of its factors is 12^3, which is 1728. He also writes the formula for the number of factors: (a+1)(b+1)(c+1) for a number N = p^a * q^b * r^c.

3. 5:00 – 10:00 05:00-10:00

   The instructor demonstrates the formula for the number of factors using the number 12. He shows the prime factorization of 12 as 2^2 * 3^1. Applying the formula, he calculates the number of factors as (2+1)(1+1) = 3 * 2 = 6. He then explains that the product of the factors of a number N is N^(number of factors / 2), which for 12 is 12^3. He also discusses the concept of infinite multiples and limited factors.

4. 10:00 – 15:00 10:00-15:00

   The instructor presents a slide titled 'Number of Factors/Divisors' and reviews the basic formula for the number of factors of a number N, which is (a+1)(b+1)(c+1) where N = p^a * q^b * r^c. He then moves to a practice problem: 'The number A4531B where A & B are single digits, the given number is divisible by 72. Then the number of factors for A + B is'. He begins to solve it by factoring 72 as 2^3 * 3^2.

5. 15:00 – 20:00 15:00-20:00

   The instructor continues solving the problem about the number A4531B. He determines that for the number to be divisible by 72, it must be divisible by both 8 and 9. He uses the divisibility rule for 8 (last three digits must be divisible by 8) to find that B must be 2. He then uses the divisibility rule for 9 (sum of digits must be divisible by 9) to find that A must be 3. He concludes that A + B = 5 and the number of factors of 5 is 2.

6. 20:00 – 25:00 20:00-25:00

   The instructor presents a new problem: 'Let N = 162, let P be the smallest number multiplied to N such that N becomes a perfect square. Let Q be the smallest number multiplied to N such that N becomes a perfect cube. Find the product of factors of P+Q'. He begins by factoring 162 as 2 * 3^4. To make it a perfect square, he needs to multiply by 2, so P = 2. To make it a perfect cube, he needs to multiply by 2^2 * 3^2, so Q = 36.

7. 25:00 – 30:00 25:00-30:00

   The instructor continues the problem with N=162. He calculates P = 2 and Q = 36, so P + Q = 38. He then needs to find the product of the factors of 38. He factors 38 as 2 * 19, so the number of factors is (1+1)(1+1) = 4. The product of the factors is 38^(4/2) = 38^2 = 1444. He then moves to the next problem.

8. 30:00 – 35:00 30:00-35:00

   The instructor presents a problem: 'x is the smallest number such that x/2 is a perfect square and x/3 is a perfect cube. Find the number of divisors of x'. He sets up the equations x/2 = a^2 and x/3 = b^3. He then combines them to get x = 2a^2 = 3b^3. He deduces that x must be divisible by 2 and 3, so he lets x = 2^a * 3^b. He then uses the conditions to find that a must be odd and b must be even, and that a-1 must be divisible by 3 and b-1 must be divisible by 2.

9. 35:00 – 40:00 35:00-40:00

   The instructor continues solving for x. He finds the smallest values for a and b that satisfy the conditions: a=3 and b=4. Therefore, x = 2^3 * 3^4 = 8 * 81 = 648. He then calculates the number of divisors of x as (3+1)(4+1) = 4 * 5 = 20.

10. 40:00 – 45:00 40:00-45:00

    The instructor presents a new problem: 'A number is divided by 406 leaves remainder 115. What will be the remainder when it will be divided by 29?'. He writes the formula Dividend = Divisor * Quotient + Remainder, so n = 406q + 115. He then divides this expression by 29 to find the remainder. He calculates 406 / 29 = 14, so 406q is divisible by 29. He then finds the remainder of 115 / 29, which is 28.

11. 45:00 – 50:00 45:00-50:00

    The instructor presents a problem: 'A number when divided by D leaves a remainder of 8 and when divided by 3D leaves a remainder of 21. What is the remainder left, when twice the number is divided by 3D?'. He sets up the equations n = Dx + 8 and n = 3Dy + 21. He equates them to get Dx + 8 = 3Dy + 21, which simplifies to D(x - 3y) = 13. He then considers the possible values for D and x-3y, finding D=13 and x-3y=1. He then finds n = 13*1 + 8 = 21, so 2n = 42. Finally, he divides 42 by 3D = 39 to get a remainder of 3.

12. 50:00 – 55:00 50:00-55:00

    The instructor presents a problem: 'If n is a natural number such that 10^100 < n < 10^101 and the sum of the digits on n is 2, then the number of values n can take is'. He explains that n is a 101-digit number. The sum of its digits is 2, so the only possibilities are a 1 followed by 100 zeros, or a 2 followed by 100 zeros, or a 1 followed by a 1 and then 99 zeros. He calculates the number of such numbers as 100 (for 1 followed by 1 and 99 zeros) + 1 (for 2 followed by 100 zeros) + 1 (for 1 followed by 100 zeros) = 102.

13. 55:00 – 60:00 55:00-60:00

    The instructor presents a problem: 'The square root of 12345678987654321 is nnnnnn... upto p times, find the sum of n and p?'. He shows that the square root of 121 is 11, the square root of 12321 is 111, and the square root of 1234321 is 1111. He identifies the pattern: the square root of a number that reads 123...n...321 is a number with n ones. The given number is 12345678987654321, which has 9 digits in the middle, so the square root is 111111111, which is 9 ones. Therefore, n=1 and p=9, so n+p=10.

14. 60:00 – 65:00 60:00-65:00

    The instructor continues the problem about the square root of 12345678987654321. He confirms that the number is a palindrome and that its square root is a number with 9 ones, which is 111111111. He then states that the sum of n and p is 1 + 9 = 10. He then moves to the next problem.

15. 65:00 – 70:00 65:00-70:00

    The instructor presents a problem: 'If n is a natural number such that 10^100 < n < 10^101 and the sum of the digits on n is 2, then the number of values n can take is'. He explains that n is a 101-digit number. The sum of its digits is 2, so the only possibilities are a 1 followed by 100 zeros, or a 2 followed by 100 zeros, or a 1 followed by a 1 and then 99 zeros. He calculates the number of such numbers as 100 (for 1 followed by 1 and 99 zeros) + 1 (for 2 followed by 100 zeros) + 1 (for 1 followed by 100 zeros) = 102.

16. 70:00 – 75:00 70:00-75:00

    The instructor continues the problem about the number of values n can take. He explains that n is a 101-digit number. The sum of its digits is 2, so the only possibilities are a 1 followed by 100 zeros, or a 2 followed by 100 zeros, or a 1 followed by a 1 and then 99 zeros. He calculates the number of such numbers as 100 (for 1 followed by 1 and 99 zeros) + 1 (for 2 followed by 100 zeros) + 1 (for 1 followed by 100 zeros) = 102.

17. 75:00 – 80:00 75:00-80:00

    The instructor presents a problem: 'The square root of 12345678987654321 is nnnnnn... upto p times, find the sum of n and p?'. He shows that the square root of 121 is 11, the square root of 12321 is 111, and the square root of 1234321 is 1111. He identifies the pattern: the square root of a number that reads 123...n...321 is a number with n ones. The given number is 12345678987654321, which has 9 digits in the middle, so the square root is 111111111, which is 9 ones. Therefore, n=1 and p=9, so n+p=10.

18. 80:00 – 81:02 80:00-81:02

    The video ends with a closing screen showing the 'KG' logo for Knowledge Gate and the website www.knowledgegate.in. The background is a dark, abstract design with glowing orange lines, signifying the end of the lecture.

This video is a structured and comprehensive lecture on the topic of number systems, specifically focusing on multiples and factors. The instructor begins with foundational definitions and formulas, using the number 12 as a primary example to illustrate the concepts of factors, multiples, and the formula for the number of factors. The core of the video is a series of practice problems that progressively increase in complexity, covering topics such as divisibility rules, finding the smallest multiplier to form a perfect square or cube, and solving remainder problems using the fundamental division algorithm. The lecture also includes a unique problem involving the square root of a palindromic number, demonstrating a pattern recognition skill. The overall teaching style is methodical, with clear step-by-step solutions, making it an effective resource for students preparing for competitive exams.