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Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant PDF Download

SUCCESSIVE DIVISION
If we divide a number by a particular number then again divide the resulting quotient by a number, and again… this is called Successive division. Let us try to understand the concept through an example.
Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant

Successive Division Method
Let us consider that 352 has to be successively divided by 6, 5, and 3. Now, to successively divide 352 the first step is to divide 352/6, which will give a quotient of 58 and a remainder of 4.

The next step is to divide the quotient obtained in this division with 5, i.e. 58/5, which will give a quotient of 11 and a remainder of 3. And now again we have to divide the quotient obtained in the last division with 3, 11/3, which will give a quotient of 3 and a remainder of 2.

So, when 352 is divided by 6, 5, and 3 successively we obtained a remainder of 4, 3, and 2 respectively. This is called successive division.

So, from the concept we saw above a question can be framed that, If a number when successively divided by 6, 5, and 3 leaves the remainders 4, 3, and 2 respectively. Find the number.

The number can be calculated as follows:
According to the concept explained below:
Step 1: Write the divisors of the division along a horizontal line.
Step 2: Write the respective remainders under the divisors.
Divisors →  6      5      3
Remainder → 4       3      2
Now, the last remainder has to be multiplied by the second divisor and then added to the second remainder, the result obtained is again multiplies by the first divisor and then added to the first remainder.
The result of the above calculation will be:
{(2 * 5) + 3} * 6 + 4 = 82
82 will be the smallest number, when successively divided by 6, 5, and 3 to leave the remainders 4, 3, and 2 respectively.

⇒ Now, the other numbers which give the same remainders when successively divided by 6,5, and 3 will be of the form.

(6 * 5 * 3) k + 82, or 90k + 82.

Step 1: Write the divisors of the division along a horizontal line.
Step 2: Write the respective remainders under the divisors.
So. the arrangement of numbers is as below, when 247 is successively divided by 7,3 and 2.
Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant

Step 3: Starting from the last remainder, multiply the number along the inclined line, then add the number along the vertical line.
Repeat the process until the first remainder is reached.
The result of the above calculation is :
[(1 × 3) + 2 ] × 7 + 2 = 5 × 7 + 2 = 37
37 is the lowest number that leaves the remainders 2, 2 and 1 when successively divided by 7 , 3 and 2.
Other numbers that have the same remainders are given by the form (7 × 3 × 2)k + 37 or 42k + 37.
Note: Where 4 = 5, value of 42k + 37 = 247, which is the number shown in the example.

The above is generalized and written as:
N = (d1, d2, d3 . . . . . . . )k + (least value).
where d1 d2, d3} etc., are the divisors of the successive division and least value is the value which is calculated as shown above.

Question for Concept of Successive division & Remainder problems on different models of HCF and LCM
Try yourself:A number when divided successively by 5, and 2 gives respective remainders of 3 and 1. What will be the remainder when the largest such two-digit number is divided by 12?
View Solution

Sample problem
1. If a number when successively divided by 7, 5, and 3 leaves the remainders 4, 2, and 1 respectively. Find the smallest such number.
(a) 42
(b) 53
(c) 67
(d) 73
Solution.
This is a problem of successive division.
Divisors   7   5   3
Remainders   4   2   1
Question for Concept of Successive division & Remainder problems on different models of HCF and LCM
Try yourself:A number when successively divided by 3, 4 and 9 leaves respective remainders of 2, 3, and 7. What will be the remainders if the original number is divided successively by 3, 5, and 7 respectively?
View Solution


REMAINDER PROBLEMS on different models of HCF and LCM

Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant

  • The greatest number that will exactly divide by x, y, and z will be the Highest common factor of x, y, and z which is HCF (x, y, z).
  • The least number that will be exactly divisible by x, y, and z will be the Least common multiple of x, y, and z which is LCM (x, y, z).
  • A number N when divided by positive integers a, b and c leaves different remainders p, q and r such that the difference between the divisors and their respective remainders are equal then the smallest value of N = LCM (a, b, c) – d.
  • A number N when divided by positive integers a, b and c leaves same remainder r in each case, then the number will be N = K * LCM (p, q, and r) + r
  • The greatest number when divided by x, y and z leaving different remainders a, b, and c respectively = HCF (x - a, y - b, z - c).
  • The greatest number when divided by x, y and z leaving the same remainder will be HCF ((x - y), (y - z), (z - x))

LCM Models

Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant 

Question for Concept of Successive division & Remainder problems on different models of HCF and LCM
Try yourself:A number when successively divided by 3, 4 and 5 leaves remainders 1, 2 and 3 respectively. Find the remainder when it would be divided by 60?
View Solution


HCF Models
Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant


Sample Problems
Q.1. What is the greatest number that will exactly divide 841, 348, and 638 leaving no remainder?
(a) 9
(b) 29
(c) 58
(d) 87
Ans. (b)
Solution.
The greatest number that will be exactly divide by 841, 348, and 638 will be    the Highest common factor of 841, 348, and 638 i.e. HCF (841, 348, and 638)

The HCF of the three numbers will be 29.
∴ 29 is the greatest number that will exactly divide 841, 348, and 638 leaving no remainder.

Question for Concept of Successive division & Remainder problems on different models of HCF and LCM
Try yourself:What is the smallest number that is exactly divisible by 39, 132, and 297?
View Solution

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FAQs on Concept of Successive division & Remainder problems on different models of HCF and LCM - Quant

1. What is the concept of successive division and remainder problems in HCF and LCM?
Ans. The concept of successive division and remainder problems in HCF (Highest Common Factor) and LCM (Least Common Multiple) involves finding the largest common factor between two or more numbers and the smallest multiple that is divisible by those numbers. Successive division is a method used to find the HCF by repeatedly dividing the larger number by the smaller number until the remainder becomes zero. The HCF is obtained when the remainder becomes zero, and the divisor at that step is the HCF. The LCM is found by multiplying the common factors and the remaining factors of the given numbers.
2. How can we solve HCF and LCM problems using the concept of successive division and remainder?
Ans. To solve HCF and LCM problems using the concept of successive division and remainder, follow these steps: 1. Write down the given numbers. 2. Choose any two numbers and perform division, dividing the larger number by the smaller number. 3. Note down the remainder obtained in each step. 4. Continue dividing until the remainder becomes zero. 5. The divisor at the step where the remainder becomes zero is the HCF of the given numbers. 6. To find the LCM, multiply the HCF by the remaining factors of the given numbers.
3. Can you explain the different models of HCF and LCM in successive division and remainder problems?
Ans. In successive division and remainder problems, there are different models for finding the HCF and LCM: 1. The Equal Difference Model: In this model, the difference between the given numbers remains the same throughout the division process until the remainder becomes zero. 2. The Co-prime Numbers Model: In this model, the given numbers are co-prime, which means they have no common factors other than 1. In such cases, the HCF is always 1, and the LCM is the product of the given numbers. 3. The Multiple Numbers Model: In this model, the given numbers are multiples of a common number. In such cases, the HCF is the common number, and the LCM is the product of the given numbers divided by the HCF.
4. What are some tips to solve successive division and remainder problems on different models of HCF and LCM?
Ans. Here are some tips to solve successive division and remainder problems on different models of HCF and LCM: 1. Identify the model of the problem: Determine if the problem falls under the equal difference model, co-prime numbers model, or multiple numbers model. This will help you understand the approach to be followed. 2. Start with smaller numbers: If the numbers are large, it is better to start with smaller numbers for division to simplify the process. 3. Keep track of remainders: Note down the remainders obtained in each step to ensure accuracy in finding the HCF and LCM. 4. Use prime factorization: If the numbers have prime factors, use prime factorization to find the HCF and LCM efficiently. 5. Practice different examples: Solve various examples from each model to gain proficiency in solving successive division and remainder problems.
5. How does understanding successive division and remainder problems in HCF and LCM help in competitive exams?
Ans. Understanding successive division and remainder problems in HCF and LCM is essential for competitive exams as these concepts often appear in mathematical aptitude sections. By mastering these concepts, you can solve such problems quickly and accurately, saving time during the exam. Additionally, knowing different models and techniques for finding the HCF and LCM helps in solving complex problems efficiently. Regular practice and understanding of these concepts can significantly improve your performance in competitive exams.
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