Concept of Reminders (Part 3)

Duration: 22 min

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This educational video is a lecture on number systems, focusing on the concept of remainders in divisibility problems. The instructor, Yash Jain, begins by introducing the topic with a title slide and then presents two complex problems. The first problem asks for the remainder when the sum of cubes from 22^3 to 88^3 is divided by 110. The second problem involves finding the sum of remainders when the sum of seventh powers from 16^7 to 19^7 is divided by 5, 7, 14, and 35. The lecture progresses by explaining the mathematical principle that if a number is divisible by two coprime numbers, it is also divisible by their product. The instructor demonstrates this by pairing terms in the first sum (22^3 + 88^3, 23^3 + 87^3, etc.) and showing that each pair is divisible by 110, thus the entire sum is divisible by 110, resulting in a remainder of 0. The second problem is solved by noting that the sum of the numbers 16, 17, 18, and 19 is 70, which is divisible by 5, 7, 14, and 35, and since the exponent 7 is odd, the sum of their seventh powers is also divisible by these numbers, leading to a sum of remainders of 0. The video uses a digital whiteboard for all explanations and includes a final thank you screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title slide for a lecture on 'NUMBER SYSTEM' by Yash Jain, a Knowledge Gate Educator. The slide features a colorful background with mathematical symbols and the tagline 'The mysterious world of numbers...'. The instructor, visible in a small window, introduces the topic, setting the stage for a lesson on number systems and remainders.

  2. 2:00 – 5:00 02:00-05:00

    The instructor presents two problems on the screen. The first is: 'If (22)^3 + (23)^3 + (24)^3 + ... + (87)^3 + (88)^3 is divided by 110, then the remainder will be __?'. The second is: 'If (16)^7 + (17)^7 + (18)^7 + (19)^7 is divided by 5, 7, 14 and 35 and remainders obtained are R1, R2, R3 and R4 respectively. Find R1 + R2 + R3 + R4.'. The instructor begins to analyze the first problem, writing out the sum and noting the range of numbers from 22 to 88.

  3. 5:00 – 10:00 05:00-10:00

    The instructor explains the strategy for the first problem. He identifies that the sum is from 22 to 88, which is an arithmetic sequence. He calculates the number of terms as 67. He then pairs the terms from the beginning and end of the sequence: (22^3 + 88^3), (23^3 + 87^3), and so on. He writes the formula for the nth term of an arithmetic sequence, an = a + (n-1)d, and calculates that 88 = 22 + (n-1)1, finding n=67. He then shows that 22+88=110, 23+87=110, etc., and notes that since the exponent 3 is odd, the sum of cubes a^n + b^n is divisible by a+b when n is odd.

  4. 10:00 – 15:00 10:00-15:00

    The instructor draws a diagram on the whiteboard to illustrate the divisibility rule for a^n + b^n. He shows that if n is odd, a^n + b^n is divisible by a+b. He writes the formula a^n + b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)). He applies this to the first problem, stating that since 22+88=110, 23+87=110, and so on, and the exponent 3 is odd, each pair (22^3 + 88^3), (23^3 + 87^3), etc., is divisible by 110. He then notes that there are 33 such pairs and one middle term, 55^3, which is also divisible by 110 because 55 is a factor of 110. Therefore, the entire sum is divisible by 110, and the remainder is 0.

  5. 15:00 – 20:00 15:00-20:00

    The instructor moves to the second problem. He writes the sum N = 16^7 + 17^7 + 18^7 + 19^7. He calculates the sum of the bases: 16+17+18+19 = 70. He notes that 70 is divisible by 5, 7, 14, and 35. He then applies the same principle: since the exponent 7 is odd, the sum of the seventh powers is divisible by the sum of the bases, which is 70. Therefore, N is divisible by 5, 7, 14, and 35, and the remainders R1, R2, R3, and R4 are all 0. The sum R1 + R2 + R3 + R4 is 0.

  6. 20:00 – 22:25 20:00-22:25

    The video concludes with a final screen. A red and blue neon-style box appears with the text 'THANK YOU FOR WATCHING'. A lightning bolt effect splits the box down the middle, and the video ends. The instructor's voice is not heard in this final segment.

The video provides a clear and structured lesson on solving remainder problems using the properties of divisibility. The core concept demonstrated is that for an odd exponent n, the sum a^n + b^n is divisible by a+b. This principle is applied to two different problems. In the first, the sum of cubes from 22 to 88 is shown to be divisible by 110 by pairing terms that sum to 110. In the second, the sum of seventh powers from 16 to 19 is shown to be divisible by 70, and since 70 is a multiple of 5, 7, 14, and 35, the sum is divisible by all of them, resulting in a sum of remainders of 0. The lecture effectively uses visual aids and step-by-step reasoning to teach a powerful mathematical shortcut.