Short Trick for Reminders on Successive Divisions

Duration: 11 min

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This educational video is a lecture on number systems, specifically focusing on the concept of remainders in division. The instructor begins by introducing the topic with a title slide and a simple example of division with a remainder, using the equation 7 ÷ 2 = 3 R1, where the remainder is highlighted. The core of the video is a problem-solving session. The first problem states that a number, when divided by 3, 4, and 7, leaves remainders of 2, 1, and 4 respectively, and asks for the remainder when the same number is divided by 84. The instructor methodically breaks down the problem using the formula: Dividend = (Divisor × Quotient) + Remainder. He sets up a table with the divisors and their corresponding remainders, then uses a step-by-step substitution method to express the original number N in terms of a variable x. He first expresses N as 3(4(7x + 4) + 1) + 2, which simplifies to N = 84x + 53. This shows that the number is of the form 84x + 53, meaning when divided by 84, the remainder is 53. The second problem involves two numbers, P and Q, with similar conditions. For P, a number is divided by 4, 5, and 7, leaving remainders 1, 4, and 1 respectively. For Q, a number is divided by 5, 6, and 8, leaving remainders 2, 3, and 4 respectively. The question asks for a possible value of P + Q. The instructor applies the same method to find that P = 140a + 53 and Q = 120b + 53. He then evaluates the given options (267, 529, 1007, 1961) by subtracting 53 from each, checking if the result is divisible by 140 and 120. He finds that 529 - 53 = 476, which is not divisible by 140, but 1007 - 53 = 954, which is not divisible by 120. He then correctly identifies that 529 - 53 = 476 is divisible by 140, and 1007 - 53 = 954 is divisible by 120, but the correct answer is 529, which is 140*3 + 53. The video concludes with a thank you message.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title slide that reads "NUMBER SYSTEM" in a stylized font against a gray background. This transitions to a colorful title slide with the text "NUMBER SYSTEM" and the subtitle "The mysterious world of numbers...". The slide also features the text "by YASH [KG]" and "YASH JAIN". The instructor, a man with glasses, appears in a small window in the bottom right corner. The video then moves to a new slide titled "Concept of Remainders". This slide shows a diagram of two puppies and bones, illustrating the concept of division. The equation 7 ÷ 2 = 3 R1 is written on the screen, with an arrow pointing to the 'R1' and the word "Remainder" written below it. The instructor explains that the remainder is the part that is left over after division.

  2. 2:00 – 5:00 02:00-05:00

    The video presents a problem on a slide with a yellow background. The text reads: "Q: On dividing a number successively by 3, 4 and 7, the remainder are 2, 1 and 4 respectively. If the same number is divided by 84, the remainder is __". The instructor begins to solve this problem by writing the formula for division: "Dividend = (Divisor × Quotient) + Remainder". He then sets up a table with the divisors (3, 4, 7) and their corresponding remainders (2, 1, 4). He starts by expressing the number N as N = 3q1 + 2, where q1 is the quotient. He then expresses q1 as q1 = 4q2 + 1, and q2 as q2 = 7q3 + 4. He substitutes these into the original equation to get N = 3(4(7q3 + 4) + 1) + 2.

  3. 5:00 – 10:00 05:00-10:00

    The instructor continues to simplify the expression for N. He expands the equation N = 3(4(7q3 + 4) + 1) + 2 to get N = 3(28q3 + 16 + 1) + 2, which simplifies to N = 3(28q3 + 17) + 2. Further simplification gives N = 84q3 + 51 + 2, resulting in N = 84q3 + 53. He explains that this means the number N is of the form 84x + 53, so when N is divided by 84, the remainder is 53. The video then transitions to a new problem. The slide text reads: "Q: There are two finite control units, in the first FCU, A number is successively divided 3 times by 4, 5 & 7 respectively leaving remainders as 1, 4, 1 respectively. The remainder comes out to be some value P when the same number is divided by 84. In the second FCU, some another number Q is successively divided 3 times by 5, 6 & 8 respectively leaving remainders as 2, 3, 4 respectively. Out of the given options, what could be the possible value of P + Q." The options are a) 267, b) 529, c) 1007, d) 1961. The instructor begins to solve for P by setting up the equation N = 4(5(7x + 1) + 4) + 1, which simplifies to N = 140x + 53. He then solves for Q by setting up the equation N = 5(6(8y + 4) + 3) + 2, which simplifies to N = 240y + 122. He then checks the options by subtracting 53 from each and seeing if the result is divisible by 140 and 120. He finds that 529 - 53 = 476, which is divisible by 140, and 1007 - 53 = 954, which is divisible by 120. He concludes that the correct answer is 529.

  4. 10:00 – 11:12 10:00-11:12

    The video concludes with a black screen featuring a red neon-style rectangle with the text "THANK YOU FOR WATCHING". A white lightning bolt effect runs vertically through the rectangle. The instructor's voice is heard saying "Thank you for watching". The video ends with this thank you message.

The video provides a comprehensive tutorial on solving remainder problems using a systematic, step-by-step approach. It begins with a foundational concept of remainders, illustrated with a simple example, and then progresses to more complex problems involving successive division. The core method demonstrated is the use of the division formula (Dividend = Divisor × Quotient + Remainder) to express the unknown number in terms of a variable. By substituting the conditions of successive division, the number is expressed in a form that reveals its remainder when divided by a specific number, often the least common multiple of the divisors. The video effectively applies this method to two distinct problems, one with a single unknown and another with two unknowns (P and Q), and uses the given options to verify the solution. The key takeaway is the power of algebraic substitution to solve problems that would be difficult to solve by trial and error.