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NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers - Exercise 3.5

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NCERT Solutions for Class 6 Maths Chapter 3 (Ex 3.5)

NCERT Solutions for Maths Class 6 Chapter 3 are provided by Vedantu to enhance your preparation for exams. To understand the basic formula and their applications, it is necessary to use solved examples and solution banks for reference, especially at the time of self-study. Maths is a subject that a lot of students find difficult to grasp. The experts at Vedantu have made a solution bank which has every step in the problem-solving process for you to understand exactly how they came to the required solution. 


Class:

NCERT Solutions for Class 6

Subject:

Class 6 Maths

Chapter Name:

Chapter 3 - Playing with Numbers

Exercise:

Exercise - 3.5

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English,Hindi will become easy to study if you have access to NCERT Solution for Class 6 Science , Maths solutions and solutions of other subjects. You can also download NCERT Solutions for Class 6 Maths to help you to revise complete syllabus and score more marks in your examinations.

Access NCERT Solutions for Class 6 Maths Chapter 3 – Playing with Numbers

Exercise 3.5

Refer to page 14 for Exercise 3.5 in the PDF

1. Which of the following statements are true:

(a) If a number is divisible by \[3\], it must be divisible by \[9\].

Ans: A number divisible by \[3\]need not be divisible by \[9\]. For example ,\[6\] is divisible by \[3\], but \[6\] is not divisible by \[9\].

Hence, the given statement is not true.


(b) If a number is divisible by \[9\], it must be divisible by \[3\].

Ans:\[3\] is a factor of \[9\].

Under the property ‘If a number a number is divisible by a number ‘a’, then it is divisible by ‘a’ factors’, a number divisible by \[9\] is also divisible by \[3\] .

Hence, the given statement is true.


(c) A number is divisible by \[18\] , if it is divisible by both \[3\] and \[6\].

Ans: \[3\]and \[6\]are factors of \[18\]. Therefore under corollary of property ‘If a number a number is divisible by a number ‘a’, then it is divisible by ‘a’ factors’ , a number is divisible by \[18\]  if it is divisible by \[3\] and \[6\].

Hence, the given statement is true.


(d) If a number is divisible by \[9\] and \[10\]both ,then it must be divisible by \[90\].

Ans: Factors of \[9\]are \[ = 1,3,9\]

         Factors of \[10\]are \[ = 1,2,5,10\]

The common factor between \[9\]and \[10\] is \[1\] , therefore \[9\] and \[10\]are co-prime numbers.

Therefore, under the property ‘If a number is divisible by co-prime numbers, then that number is also divisible by product of those co-prime numbers’, a number divisible by \[9\]and \[10\] is also divisible by 

Hence, the given statement is true.


(e) If two numbers are co-primes, at least one of them must be prime.

Ans: The given statement is not true. For example, \[9\]and \[10\]are co-primes, but \[9\] and\[10\] are not a prime number.

Hence, the given statement is not true.


(f) All numbers which are divisible by \[4\] must also be divisible by \[8\].

Ans: A number divisible by \[4\]need not be divisible by \[8\]. For example ,\[12\] is divisible by\[4\], but \[12\] is not divisible by \[8\].

Hence, the given statement is not true.


(g) All numbers which are divisible by \[8\] must be divisible by \[4\].

Ans: \[4\]is a factor of \[8\].

Under the property,If a number is divisible by a number ‘a’, then it is divisible by ‘a’ factors’, a number divisible by \[8\]is also divisible by \[4\].

Hence, the given statement is true.


(h) If a number exactly divides two numbers separately, it must exactly divide their sum.

Ans: Under the property, ‘if  two numbers are divisible by a number, then their sum is also divisible by that number’ , the given statement is true.


(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

Ans: Under the corollary of the property, ‘if two numbers are divisible by a number, then their sum is also divisible by that number’ , the given statement is true.


2) Here are two different factors trees for \[60\]. Write the missing numbers.

(a)


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Ans: Let’s first consider the number missing in left branch:

Prime factorisation of \[6\] is \[2 \cdot 3\]. \[2\] is given but \[3\]is missing , therefore number missing in left branch is \[3\] . 

Now let’s consider the number missing in right branch:

Prime factorisation of \[10\] is \[2 \cdot 5\]. \[5\] is given but \[2\] is missing in the right branch, therefore number missing in right branch is \[2\] . 


(b)


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Ans: Let consider the first branch:

Factor pair of \[60\] which includes \[30\]  is \[60 = 2 \cdot 30\] . Because \[30\] is given but \[2\]is missing , therefore number missing in the first branch is \[2\].

Now let’s consider the second branch:

Factor pair of \[30\] which includes \[10\]is \[30 = 3 \cdot 10\] . Because \[10\]is given but \[3\] is missing , therefore number missing in the second branch is \[3\].

Now let’s consider the third branch:

Prime factorization of \[10\] is \[2 \cdot 5\] therefore \[2\]and \[5\]are the missing numbers in the third branch.


3. Which factors are not included in the prime factorization of a composite number?

Ans:  Composite number is not a prime number because composite number is a number with more than two factors, i.e., \[1\]and the number itself. \[1\] is not a prime number. Because prime factorisation involves only prime factors, \[1\] and the composite number itself are not included in the prime factorisation of a composite number.


4. Write the greatest 4-digit number and express it in terms of its prime numbers.

Ans: Greatest 4- digit number is \[9999\].

We divide the number \[9999\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.

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Therefore  \[9999\]  in terms of prime factors is \[{3^2} \cdot 11 \cdot 101\]


5. Write smallest 5-digit number and express it in the form of its prime factors.

Ans: Smallest 5- digit number is\[10000\].

We divide the number \[10000\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.

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Therefore, \[10000\] in terms of prime factors is \[{2^4} \cdot {5^4}\].


6. Find all the prime factors of \[1729\] and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Ans: We divide the number \[1729\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.

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Therefore, \[1729\] in terms of prime factors is \[7 \cdot 13 \cdot 19\].

Prime factors of \[1729\]in ascending order are \[7,13,19\].

One can observe that \[13 - 7 = 19 - 13 = 6\],i.e., difference between two consecutive prime factors is \[6\]. Therefore, relation between two consecutive prime factors of \[1729\]is that difference between them is a constant, \[6\].

7. The product of three consecutive numbers is always divisible by \[6\] . Verify this statement with help of some examples.

Ans: Consider \[4,5\] and \[6\]. Product of\[4,5\] and \[6\] is \[120\].

\[\frac{{120}}{6} = 20\] ,i.e., \[120\]is divisible by 6.

Consider \[5,6\]and \[7\]. Product of\[5,6\]and \[7\]is \[210\].

\[\frac{{210}}{6} = 35\] , i.e., \[210\]is divisible by 6


8. The sum of two consecutive odd numbers is divisible by \[4\]. Verify this statement with the help of some examples.

Ans: Consider \[3\]and\[5\]. The sum of   \[3\]and\[5\]is \[8\].

\[\frac{8}{4} = 2\],i.e., \[8\]is divisible by \[4\].

Consider \[7\]and \[9\] . The sum of \[7\]and \[9\]  is \[16\].

\[\frac{{16}}{8} = 2\],i.e., \[16\]is divisible by \[4\].


9. In which of the following expressions, prime factorisation has been done?

(a)  \[24 = 2 \cdot 3 \cdot 4\]

Ans: In the given expression\[4\]is not factorised further into its prime factors ,i.e,. Therefore, in the given expression, prime factorisation has not been done.


(b) \[56 = 7 \cdot 2 \cdot 2 \cdot 2\]

Ans: In the given expression, all factors are prime, therefore in the expression prime factorisation has been done.


(c) \[70 = 2 \cdot 5 \cdot 7\]

Ans: In the given expression, all factors are prime , therefore in the expression prime factorisation has been done.


(d) \[54 = 2 \cdot 3 \cdot 9\]

Ans: In the given expression \[9\]is not factorised further into its prime factors ,i.e,\[3 \cdot 3\]. Therefore in the given expression, prime factorisation has not been done.


10. Determine if \[25110\]is divisible by  \[45\].

Ans: If a number is divisible by co-prime numbers then it is also divisible by their product.

Co-prime factors pair  of \[45\]is\[5\] and \[9\].

Therefore \[25110\]is divisible by \[45\] if it is divisible by \[5\]and \[9\].

\[\frac{{25110}}{5} = 5022\] and \[\frac{{25110}}{9} = 2790\],i.e., \[25110\]is divisible by \[5\]and \[9\].

Thus, \[25110\]is divisible by \[45\].


11. \[18\] is divisible by both \[2\] and \[3\]. It is also divisible by \[2 \cdot 3 = 6\]. Similarly, a number is divisible by both \[4\]and\[6\]. Can we say that the number must also be divisible by \[4 \cdot 6 = 24\]? If not, give an example to justify your answer.

Ans: No, we can not say that if a number is divisible by both \[4\]and \[6\]must  be divisible by \[4 \cdot 6 = 24\].

For example, consider \[12\] :

\[12\]is divisible by both \[4\]and \[6\]but is not divisible by \[24\].

12. I am the smallest number, having four different prime factors. Can you find me?

Ans: The smallest number with four different prime factors must be the product of first four prime numbers ,i.e., \[2,3,5\]and \[7\].

Therefore smallest number with four different prime factors is :

\[2 \cdot 3 \cdot 5 \cdot 7 = 210\]

Chapter 3: Playing with Numbers (14-15)

Introduction:

Playing with Numbers is a chapter in NCERT Maths Class 6 wherein several numerical tricks are taught to make the problem solving easier and quicker. It is extremely important to master all these numerical tricks and rules so that they can be used to solve the mathematical problem of higher-level efficiently. This lesson acts as a foundation to the whole subject and is repeated in Class 8th with some more complicated numerical tricks. This chapter is of importance as it forms the basis for a similar chapter covered in Class 8th. While mathematics is an application-based subject in every possible way, this chapter forms a very basic method for mathematical research. It covers a lot of topics, and each trick is given to use in the exercise that follows the same. The chapter covers topics such as the relationship between factors and multiples, Prime and Composite numbers, Tests of divisibility of numbers 10, 5, 2, 3, 6, 4, 8, 9, 11; common factors and multiples, prime factorisation, Highest common factor, Lowest common multiple, etc. after covering each of these topics, the chapter has exercises and solved examples so that the students can understand how the trick works in solving mathematical problems. We are going to concentrate on exercise 3.5 individually. The solution to exercise 3.5 of Class 6 Maths Chapter 3 can be downloaded from the Vedantu website. The site has PDF documents available for download for each exercise in this chapter individually. Before we move on to the subtopics covered in this chapter, let us try to understand the subject, this chapter comes under- Arithmetic.

Arithmetic: Arithmetic is a branch of mathematics that has the study of numbers. This includes the properties of various numbers; the various operations performed on them like addition subtraction multiplication and division. It is the very basic idea of the subject mathematics and forms a foundation for all ideas related to the subject. Arithmetic is also known as the number theory. Man has been using this field of study for years. Right from the prehistoric period through the first documented use of calculations in Egypt and Roman Empires, we can observe the growth of this subject. India has produced great minds in this field for many years. The world as a whole has contributed to the growth and development of arithmetic through various theories and formulas used to determine numbers and calculate results.

Besides the basic operation of addition, subtraction, multiplication and division arithmetic has multiple other operations like square root, percentage, exponential, trigonometric functions and many more. The various expressions in arithmetic are evaluated according to the intended sequence of operations. The most useful method to specify this is known as infix notation. Let us look at the basic operations in the subject.

  • Addition: addition is the combination of two or more numbers to form a third number. A result is a number greater than any number used in operation. The addition is used to calculate the result of a combination of different numbers. It is the most common mathematical operation and is taught immediately after numbers in pre-school. 

  • Subtraction: subtraction is the difference between the two numbers. If the number from which another number is being subtracted is greater than the first number, the answer will be in positive integers, and if not, then the answer will be in negative integers.

  • Multiplication: multiplication combines two numbers into a third known as the product. The two numbers are the multiplier and multiplicand, also known as factors of the resulting number. Multiplication is associative and cumulative. Any number multiplied by it is the square of the number. When any number is multiplied by 1 it results in the number itself. Any number multiplied by 0 is 0 — the multiplication of any positive integer results in a number which is larger than the two factors.

  • Division: Division is the exact opposite of multiplication. In division, a number is separated into equal parts. There is a quotient which is several parts in which the number is being divided. The dividend is the number and divisor is the number devising the dividend. Any number divided by 0 is undefined. When any number is divided by one, it results in the number itself. When any number is divided by itself, it results in 1.

Besides these four basic operations, arithmetic is divided into various topics which are covered through the years of study and are applied everywhere. There is decimal arithmetic, compound unit arithmetic number theory and many more. It is an interesting field of study and research.

A minuscule part of the whole part us covered in Maths NCERT Solutions Class 6 chapter 3 exercise 3.5. But the various facts and tricks are important all the same.

Topics covered in this chapter are:

  • Factors and multiples

  • Prime and composite numbers

  • Tests of divisibility of numbers

  • Common factors and common multiples

  • Some more divisibility rules

  • Prime factorisation

  • Highest common factor

  • Lowest common multiple

  • Some problems of HCF and LCM

Factors and Multiples:

This is an important segment of NCERT Solutions for Class 6 Maths Chapter 3 exercise 3.5. A number is said to be a factor of another when it can be divided by the first number completely without a remainder. If a number is exactly divisible by another number, the second is said to be the factor if the first. If a dividend is exactly divisible by the divisor, then the divisor is a factor of the dividend.  A multiple is a product of some given numbers. It is the result of the multiplication of two numbers. We explore these two terms in this subtopic and the various rules of mathematics that they follow. 

Let us take an example-

4*2=8 here 4 and 2 are the factors of 8 and 8 is the multiple of 4 and 2.

The subtopic in NCERT Maths Class 6 chapter 3 exercise 3.5 deals with numerous facts about factors and multiples. With solved examples and exercises that follow, we can understand the following. 1 is a factor of every number. We can also see that every number is a factor of itself too, and every factor of a number is the exact divisor of that number. Further, the subtopic explains that every factor of any number is lower than or equal to the said number.

Moreover, every multiple of any given number is more than or equal to the same number. Take into consideration the multiples of any possible number. We realise that the number of multiples of any number is infinite. We also realise that every number is a multiple of itself. The subtopic also explores the term known as a perfect number. A perfect number is a number whose sum of all factors is equal to twice the number. 6 and 28 are perfect numbers.  In the exercise that follows this subtopic, we explore the same rules and facts given above through application. Vedantu has solved examples which you can refer to on this subtopic of this chapter. These solved examples are done by various possible methods so that you can determine which method suits you the best.

Application of factors in real life: factors are extremely useful for several things. They are used to divide something into equal parts; they are used in the banking sector for factoring the money, and they are also used for comparing prices and understanding time which is the most basic use of factors.

Prime and Composite Numbers:

Prime numbers were created around 200 BC by Eratosthenes. He created an algorithm that calculated the prime numbers in the number series. Prime numbers are the foundation of whole numbers. There are multiple ways by which prime numbers can be represented visually. They are important elements in the factorisation process.

Any whole number which is divisible by only 1 and itself is known as a prime number. We can say that a prime number has just two factors. A composite number has multiple factors. 1 and 0 are not prime numbers, nor are they composite numbers. All whole numbers are divided into prime and composite numbers. All whole numbers are also divided into odd and even numbers. 

When one is added to a multiple of 2, we get a prime number. All multiples of 2, 3, 5, 7, 9 are composite numbers. The term even numbers are given to the multiples of 2.  The rest of the numbers which aren’t multiples of two are called odd numbers. Two is the smallest prime even number. Except 2 every prime number is an odd number. The term composite number was coined by an Indian mathematician called Ramanujan in 1915.

Prime numbers up to 50 are: 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. 

The rest are known as composite numbers up to 50. The distinction between prime and composite numbers is necessary as it becomes useful to solve higher-level multiplication and division problems. Students are required to learn the prime numbers at least up to 100 for better results. The exercise that follows this subtopic deals with the application and uses of these divisions of numbers while solving mathematical problems. Vedantu has solutions to the exercise that follows this subtopic. These solutions of Class 6 Maths Chapter 3 exercise 3.5 are put together by experts in the field of Maths and are very useful.

Application of prime numbers in real life: prime numbers are used in asymmetric cryptography. They are also used in Engineering along with co primes to avoid resonance and to ensure that the cogwheels wear equally.

Test of Divisibility of Numbers:

Test of divisibility of a number is a rule to decide whether a number is divisible by some other number. The history of divisibility of numbers goes back to 500 CE. At this point, the divisibility test for 7 was written in the Babylonian Talmud. From that time, multiple methods of divisibility of positive whole numbers have been discovered and lost to be discovered again. A summary of the divisibility tests is written in History of the theory of numbers by Leonard Dickson. Some numbers have some factors by which we can determine their divisibility at a glance. This chapter covers such tests of divisibility of a few basic numbers.

  • Divisibility by the number 10: If a number has 0 in its units place we can say that the number is divisible by 10. For example, numbers such as 20, 3000, 40, 60 are all divisible by 10 that is they can be divided by 10.

  • Divisibility by the number 5: If a number has 5 or 0 in its units place it can be divided by 5. All numbers that are divisible by 10 are always divisible by 5. For example, 55, 70, 65, 555, 6500 are all divisible by 5.

  • Divisibility by the number 2: Any number with 2, 4, 6, 8 and 0 at its units place is divisible by 2. It can be observed that all even numbers are divisible by 2. For example, 632, 666, 85552 are numbers divisible by 2.

  • Divisibility by the number 3: if the sum of the digits in a number is a multiple of 3 then the number is divisible by 3. For example, 54. 5+4=9. 9 is a multiple of 3. Thus, 54 is divisible by 3. 

  • Divisibility by the number 6: any number which is divisible of both 2 and 3 is divisible by 6. This means the number should be an even number whose sum of all digits is a multiple of 3. For example 24, 60, 312 etc.

  • Divisibility by the number 4: any number which has 3 or more than 3 digits is divisible by 4 if the number formed by the last two numbers can be divided by 4. For example, 222324. In this number, the number formed by the last two digits is 24, which is divisible by 4 thus the number itself is divisible by 4.

  • Divisibility by the number 8: when a number has more than 4 digits it is divisible by 8 only if the number made by the last three numbers can be divided by 8 for example 2104. The number formed by the last three digits is 104 which is divisible by 8, and thus the number 2104 can be divided by 8.

  • Divisibility by the number 9: divisibility test of 9 is similar to that of 3. If the sum of all digits in a number is divisible by 9 then the number is divisible by 9. For example, 4608. 4+6+0+8= 18. 18 is a multiple of 9, which is how we can determine that 4608 is divisible by 9.

  • Divisibility by the number 11: The number can be divided by 11 if the difference in the sum of digits at the odd placements from right and sum of digits at even placements from right is a multiple of 11.

All these tools are immediately applied in the exercise that follows in this chapter. This is one of the most important parts of mathematics and is used multiple times throughout the syllabus in higher Classes. You should refer to the Vedantu website where you can check the application of this and more divisibility tests. Learning about the divisibility test makes problem-solving in the later years much easier.

Application of divisibility tests in real life: Divisibility tests can be used to determine the value of an object in comparison to the other; basically it can be used for a comparative study. They are used for efficiency and speed in solving mathematical problems that occur in real life. They are used to determine the division of cost, and they are used in the banking sector more often than not. They are also used by engineers’ especially civil engineers and architects.

Common Factors and Common Multiples:

When a number is a multiple of two or more numbers, then it is known as a common multiple of those numbers. When two numbers can be multiplied together to get the third number, they are known as the common factors of that number. The term co-prime numbers are used for numbers which have only 1 as a common factor between them. As the name suggests, these are prime numbers. 4 and 15 are co-prime numbers. This topic is covered in detail on the Vedantu website. You should refer to the same to have a deeper understanding of the subject.

Some More Rules of Divisibility:

This subtopic of NCERT Solutions for Class 6 Maths Chapter 3 ex 3.5 adds to the divisibility rules already covered earlier in this chapter. These rules help solve mathematical problems quickly and efficiently. The rules mentioned are as follows. 

  • If any number can be divided by some other number, then it can be divided by every factor of that same number. This means that all factors of the number also act as multiples of that number.

  • If we take two co-prime numbers and a number is divisible by them, that number is also divisible by their product. This rule says that the product of two co-prime numbers is also the multiple of the number whose multiples the two co-prime numbers are.

  • If two given numbers are divisible by a number, then their sum is also divisible by the number. So basically the result of adding two numbers is divisible by a number of the added numbers are divisible by the same numbers.

  • If we take two numbers and a number is divisible by them, then the number is divisible by their difference too.

The Vedantu website has solved examples which explain the steps taken and reasons for the solutions. They will help you understand this subtopic better.

Prime Factorisation:

A prime factorisation is a tool used to determine which prime numbers when multiplied form the originally given number.it was first used and determined in Greece by the Greek mathematicians in the case of integers. They were the ones who proved that every positive integer might be factored into multiple prime numbers which can be further factored into integers. This is the fundamental theorem of arithmetic. You start by dividing the given number by the smallest of all prime numbers that are 2; you keep dividing till you get a remainder or a decimal. Then you divide by 3, 5, 7 etc. then you write the number as a product of multiple prime numbers. Now you can easily lear 6th Class Maths playing with numbers exercise 3.5.

Application of prime factorisation in real life: It is most commonly used in the RSA encryption process and has been properly entailed in Class 6 maths NCERT Solutions chapter 3 exercise 3.5. This use is based on the fact that it is extremely difficult to factorise a composite number whose prime factorisation is a two-digit prime number.

The exercise following this factor is one of the most important exercises covered in this chapter. Exercise 3.5 of chapter 3 Maths Class 6 NCERT not only covers the last two rules of the lesson but also dives into the rules previously covered in the chapter. The exercise starts with a true or false and then goes to factorisation methods. This exercise covers all the rules of divisibility tests and factorisation in general with examples. The Vedantu website has this and many more solutions to exercises in this chapter. These solutions are very useful, especially for last-minute study and assignments. Prime factorisation is later used in the engineering Classes if you chose to go for engineering; it may be civil, mechanical or ENTC. This chapter is covered thoroughly on the Vedantu website, which you should refer to for more information on the subject.

Highest Common Factor:

The highest common factor, also known as the greatest common divisor is the highest number which can be divided into two or more numbers exactly. This is used in simplifying a factor. We can find the highest common factor by finding the factors of both the numbers, determining the common factors and recognising the highest. We can also find all the common prime factors to recognise the number which can be or is close to the highest common factor. Or we can play around with numbers till we get the highest common factor of the same. This is one of the most popular terms and rules covered in this lesson. You should refer to the Vedantu website to get detailed information on this topic of importance. The procedure to find the highest common factor is known as the Euclidean algorithm.  This extremely efficient procedure is still used in computer programming after a few minor modifications. It is used to reduce a common factor to its lowest terms. It is also a very useful tool by which we can find the integer solutions to linear equations. Vedantu website and e-learning application have the means and examples you can refer to solve the paper well and score well on your tests. Moreover, as this lesson is a foundation for multiple lessons in higher Classes, this lesson needs to be done well. Keeping this in mind, the Vedantu website has separate solutions for each exercise in this chapter.

Lowest Common Multiple:

The lowest possible number, which is a multiple of two or more numbers, is known as the lowest common multiple. It is also known as the least common multiple of LCM in short.  One of the most common ways to determine the lowest common multiple of a number is to find the prime factors of each number and determine the common prime factors. Find the greatest common factors of the number. Then you divide this. Take the answer and multiply it by other numbers. You should refer to Vedantu to know more about the methods used to determine the Lowest common multiple of given numbers. Unlike languages, it is quite possible to get a full score in math. Vedantu and the guidance of their experts in grasping the Maths Class 6 chapter 3 exercise 3.5 solutions can help you achieve the same.

Some Problems of HCF and LCM:

The lesson continues with some examples of determining the lowest common multiple and highest common factor of numbers. Through these examples, you can realise the various methods to find the necessary highest common factor or lowest common multiple. There are multiple ways to find out HCF and LCM.  Two of the most common are by prime factorisation method and by division method.

HCF and LCM Tricks and Formulae:

  • Product of two numbers is equal to the product of their Highest Common Factor and Least Common Multiple.

  • Highest Common Factor is equal to the Highest Common Factor of the numeration divided by the Least Common Multiple of the denominator.

  • Least Common Multiple is equal to the Least Common Multiple of the numerator divided by the Highest Common Factor of the denominator.

Besides these solved examples, Vedantu website has many more solved examples you can refer to regarding LCM and HCF. This topic is covered in utmost detail on their website and has proved to be particularly useful to go back to in the later years because of its precise methodology. Math is a scoring subject and if done well can help change the average percentage and grade of the students.

About Vedantu:

Vedantu is one of the most popular e-learning websites in India. It covers the entire syllabus of NCERT till Class 12 along with multiple sessions on all the competitive exams in academics. You can now find solutions to NCERT Maths chapter 3 online. These solutions are put together by experts in mathematics and can help the students in learning about the basics of Maths. Maths is an area where practice can make you perfect. The site has solution banks compiled by experts in the field and many tutors from the most elite colleges- IITs, IIMs etc. these experts have general, current and in-depth knowledge about their subjects and their application and give the best possible guidance to the students. This is extremely beneficial for the students and helps them score good marks in their exams. 

Vedantu has yet to have a negative review with multiple positive responses from all students as well as their parents. The Vedantu website offers various ways to help the students. They have lectures given by the subject experts covering every aspect in a subject. They have live interactive sessions with the students where the students can clear any doubts they have. They have solution banks on each exercise of every subject till Class 12. The students can download this solution banks for free from Vedantu website. Vedantu also has multiple tests through which students can understand their level for the subject and improve accordingly. Vedantu also has an application which you can download from the play store. This app is extremely useful and can be used for all the activities mentioned above and exercises. Moreover, the app has a live chat section where you can ask for guidance and get your problems solved and explained by the subject experts. This app has been useful for students as a secondary means of revision before the exams and long term revision and problem-solving. 

The experts at Vedantu have in-depth knowledge about the subject and also about how to make it easy for the students to grasp. They concentrate on the application of mathematics rather than just the formulae. This method is of great importance, especially when it comes to Maths of higher Classes. If the foundation is strong, it becomes easier to love the subject. The solutions to 6th Class Maths Playing with Numbers exercise 3.5 are available online for you to download on the Vedantu website. They are available in the PDF format for your convenience so that you can view the solutions on any device anytime and at any place you need to. This has helped a lot of students get high scores in their tests. Students who do not go to any coaching Classes have especially benefited from the solutions to chapter 3 NCERT Maths as it is an application based lesson which is ignored by or given less importance by the school teachers. These solutions which can be downloaded for free are available on the Vedantu website.

FAQs on NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers - Exercise 3.5

Q1. Which concept is used in Exercise 3.5 of Chapter 3 of Class 6 Maths?

“Playing With Numbers” is Chapter 3 of Class 6 Maths. Exercise 3.5 is based on the concept of “Some More Divisibility Rules” and “Prime Factorisation”. In this exercise, students will get to know about some more rules for visibility which will be used while solving questions. They will also learn how to do the prime factorisation of the given numbers. The real-life application of prime factorisation is that it is used in the encryption process of RSA. This exercise will help in having a better understanding of the topics.

 Q2. Express the greatest 4-digit number and smallest 5-digit number in terms of its prime factors.

(a) Greatest 4-digit number

9999 is the greatest 4-digit. The prime factorisation of this number is:

9999 = 9 × 1111

9 = 3 × 3.         1111 = 11 × 101

Hence, 9999 = 3 × 3 × 11 × 101

(b) Smallest 5-digit number

10000 is the smallest 5-digit number

10000 = 2500 × 4

4 = 2 × 2.          2500 = 25 × 100

25 = 5 × 5.        100 = 20 × 5

20 = 4 × 5.        4 = 2 × 2

Therefore, 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Q3. Show whether 25110 is divisible by 45.

For showing the divisibility of 25110 by 45, we first need to show the divisibility of 25110 by 5 and 9.

45 = 9 × 5

Factors of 9 are 1, 3 and 9.

Factors of 5 are 1 and 5.

Hence, 5 and 9 are co-prime numbers.

Here, we can see that 0 is the last digit of 25110. Therefore, it is divisible by 5.

25110 = 2 + 5 + 1 + 1 + 0 = 9

As the sum of the digits of the given number is 9, hence it is also divisible by 9.

Since the number 25110 is divisible both by 5 and 9. Hence, it is divisible by 45.

Q4. Where can I get the NCERT Solutions of Exercise 3.5 of Chapter 3 of Class 6 Maths?

The students will find the NCERT Solutions of Exercise 3.5 of Chapter 3 of Class 6 Maths on the website of Vedantu. Beneath are steps for downloading these NCERT Solutions:

  •      Click on the link provided.

  •        The link will take you to the official page of Vedantu.

  •        There you will see the PDF file of the NCERT Solutions.

  •        At the peak of this PDF file, you will find the “Download PDF” option.

  •        Hit that option to download the NCERT Solutions in PDF format.

The solutions provided by Vedantu are free of cost. The solutions are available in both Vedantu website as well as Vedantu Mobile app.

Q5. How can I make a better study plan for studying Exercise 3.5 of Chapter 3 of Class 6 Maths?

Beneath are the points that should be kept in mind while making an effective study plan for preparing Exercise 3.5 of Chapter 3 of Class 6 Maths:

  •  Construct a timetable so that you can get time to study Maths subject.

  • Prefer the NCERT book to study Exercise 3.5 of Chapter 3 Class 6 Maths.

  • Solve each example and question of the exercise to understand the theory              clearly.

  • Make separate registers for the formulas.

  • Avert studying for longer periods.

  • Refresh your mind so that you can concentrate on your studies.