Class 10 Mathematics

Real Number

Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorised ) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

This theorem also says that the prime factorisation of a natural number is unique, except for the order of its factors.

For example 20 can be expressed as `2xx2xx5`

Using this theorem the LCM and HCF of the given pair of positive integers can be calculated.

LCM = Product of the greatest power of each prime factor, involved in the numbers.

HCF = Product of the smallest power of each common prime factor in the numbers.


Exercise 1.2 (NCERT)

Question 1: Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution:

  1. `140=2xx2xx5xx7=2^2xx5xx7`
  2. `156=2xx2xx3xx13=2^2xx3xx13`
  3. `3825=3xx3xx5xx5xx17=3^2xx5^2xx17`
  4. `5005=5xx7xx11xx13`
  5. `7429=17xx19xx23`

Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54


(i) 26 and 91

Solution: The prime factors of `26 = 2 xx 13`

The prime factors of `91 = 7 xx 13`

Therefore, LCM `= 2xx 7 xx 13 = 182`

And, HCF = 13

Now, `text(LCM) xx \text(HCF) = 182 xx 13 = 2366`

Product of given numbers `= 26 xx 91 = 2366`

Therefore, LCM × HCF = Product of the given two numbers

(ii) 510 and 92

Solution: The prime factors of `510 = 2 xx 3 xx 5 xx 17`

The prime factors of `92 = 2 xx 2 xx 23`

Therefore, LCM `= 2 xx 2 xx 3 xx 5 xx17 xx 23 = 23460`

And, HCF = 2

Now, LCM × HCF `= 23460 xx 2 = 46920`

Product of given two Numbers `= 510 xx 92 = 46920`

Therefore, LCM × HCF = Product of given two numbers

(iii) 336 and 54

The prime factors of `336 = 2 xx 2 xx 2 xx 2 xx 3 xx 7 = 2^4 xx 3 xx 7`

The prime factors of `54 = 2 xx 3 xx 3 xx 3 = 2 xx 3^3`

Therefore, LCM of 336 and 54 `= 2^4 xx 3^3 xx 7 = 3024`

And, HCF `= 2 xx 3 = 6`

Now, `text(LCM) xx \text(HCF) = 3024 xx 6 = 18144`

And the product of given numbers `= 336 xx 54 = 18144`

Therefore, LCM × HCF = Product of given numbers

Question 3: Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

(i) 12, 15 and 21

Solution: Prime factors of `12 = 2 xx 2 xx 3 = 2^2 xx 3`

Prime factors of `15 = 3 xx 5`

Prime factors of `21 = 3 xx 7`

Therefore, LCM `= 2 xx 2 xx 3 xx 5 xx 7 = 420`

HCF = 3

(ii) 17, 23 and 29

Solution: Prime factors of `17 = 17 xx 1`

Prime factors of `23 = 23 xx 1`

Prime factors of `29 = 29 xx 1`

Therefore, LCM `= 17 xx 23 xx 29 = 11339`

And HCF = 1

(iii) 8, 9 and 25

Prime factors of `8 = 2 xx 2 xx 2 = 2^3`

Prime factors of `9 = 3 xx 3 = 3^2`

Prime factors of `25 = 5 xx 5= 5^2`

Therefore, LCM `= 2^3 xx 3^2 xx 5^2 = 8 xx 9 xx 25 = 1800`

Since, there is no common factors among the prime factors of given three numbers,

Thus HCF = 1


Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution: We know that `text(LCM)xx\text(HCF)=text(Product of given numbers)`

Or, `text(LCM)=text(Product of number)/text(HCF)`

Or, `text(LCM)=(306xx657)/(9)`

`=(201042)/(9)=22338`

Question 5: Check whether 6n can end with the digit 0 for any natural number n.

Solution: Numbers that ends with zero are divisible by 5 and 10. Simultaneously they are divisible by 2 and 5 both.

If the number 6n will be divisible by 2 and 5, then, it will end with the digit 0 otherwise not.

Prime factor of `6n = (2 xx 3)^n`

Since, 5 is not a prime factor of 6n

Therefore, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

Note: 6n always has 6 at unit’s place. Following examples illustrate this:

`6^1 = 6`

`6^2 = 36`

`6^3 = 216`

`6^4 = 1296`

Hence, 6n can never end with the digit zero for any natural number n.

Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution: Numbers which have at least one factor other than 1 and number itself are called composite numbers.

Given, first expression:

`7 xx 11 xx 13 + 13`

`= 13 xx (7 xx 11 + 1)`

`= 13 xx (77 + 1)`

`= 13 xx 78`

Since, given expression `7 xx 11 xx 13 + 13` has two prime factors other than 1, thus it is a composite number.

Second expression:

`7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1 + 5`

`= 5(7 xx 6 xx 4 xx 3 xx 2 xx 1 + 1)`

`= 5(1008 + 1)`

`= 5 xx 1009`

Since, the given expression has two factors other than 1, thus it is a composite number.

Question 7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: Since, they will take different times to complete one round of the sports field. Thus, time after which they will meet again at the starting point will be given by the LCM of time taken to complete one round for each of them.

Time taken by Sonia to complete one round = 18 minute

Time taken by Ravi to complete one round = 12 minute

Prime factors of `18 = 2 xx 3 xx 3 = 2 xx 3^2`

Prime factors of `12 = 2 xx 2 xx 3 = 2^2 xx 3`

Therefore, LCM `= 2^2 xx 3^2 = 4 × 9 = 36`

Thus, after starting simultaneously Ravi and Sonia will meet at starting point after 36 minute.


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Class ten Math - Real Numbers - Solution of NCERT Exercise 1.1

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