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**Exercise 1.2Ques 1: Express each number as a product of its prime factors:(i) 140 **

âˆ´ 140 = 2 Ã— 2 Ã— 5 Ã— 7

= 2^{2} Ã— 5 Ã— 7 **(ii) Using factor tree method, we have:**

(ii) 156 = 2 x 2 x 3 x 13

= 2^{2} x 3 x 13**(iii) Using factor tree method, we have:**

3825 = 3 Ã— 3 Ã— 5 Ã— 5 Ã— 17

= 3^{2} Ã— 5^{2} Ã— 17**(iv) Using factor tree method, we have:**

âˆ´ 5005 = 5 Ã— 7 Ã— 11 Ã— 13**(v) Using the factor tree method, we have:**

âˆ´ 7429 = 17 Ã— 19 Ã— 23**Ques 2: Find the LCM and HCF of the following pairs of integers and verify that LCM Ã— HCF = product of the two numbers.(i) 26 and 91 **

Sol: (i) Using factor-tree method, we have:

âˆ´ LCM of 26 and 91 = 2 Ã— 7 Ã— 13 = 182

HCF of 26 and 91 = 13

Now, LCM Ã— HCF = 182 Ã— 13 = 2366

and 26 Ã— 91 = 2366

i.e., LCM Ã— HCF = Product of two numbers.**(ii) Using the factor tree method, we have:**

âˆ´ LCM of 510 and 92 = 2 Ã— 3 Ã— 5 Ã— 17 Ã— 2 Ã— 23

= 23460

HCF of 510 and 92 = 2

â‡’ LCM Ã— HCF = 23460 Ã— 2 = 46920

510 Ã— 92 = 46920

i.e., LCM Ã— HCF = Product of two numbers.**(iii) Using the factor tree method, we have:**

LCM of 336 and 54 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 7 = 3024

HCF of 336 and 54 = 2 Ã— 3 = 6

LCM Ã— HCF = 3024 Ã— 6 = 18144

Also 336 Ã— 54 = 18144

Thus, LCM Ã— HCF = Product of the numbers**Ques 3: Find the LCM and HCF of the following integers by applying the prime factorisation method.****(i) 12, 15 and 21 ****(ii) 17, 23 and 29 ****(iii) 8, 9 and 25****Sol. ****(i) We have**

âˆ´ HCF of 12, 15 and 21 is 3

LCM of 12, 15 and 21 is 2 Ã— 2 Ã— 3 Ã— 5 Ã— 7 or 420**(ii) We have:**

17 = 1 Ã— 17 | â‡’ HCF of 17, 23 and 29 is 1 |

23 = 1 Ã— 23 | LCM of 17, 23 and 29 |

29 = 1 Ã— 29 | = 17 Ã— 23 Ã— 29 = 11339 |

**(iii) We have:**

âˆ´ HCF of 8, 9 and 25 is 1

LCM of 8, 9 and 25 is 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5 or 1800**Ques 4: Given that HCF (306, 657) = 9, find LCM (306, 657).****Sol: **Here, HCF of 306 and 657 = 9

We have:

LCM Ã— HCF = Product of the numbers

âˆ´ LCM Ã— 9 = 306 Ã— 657

â‡’ LCM = 34 Ã— 657/9

= 22338

Thus, LCM of 306 and 657 is 22338.**Ques 5: Check whether 6 ^{n} can end with the digit 0 for any natural number n.**

âˆ´ 6

But the prime factors of 6 are 2 and 3.

â‡’ 6

i.e., In the prime factorisation of 6

Since, by the Fundamental Theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorisation occurs.

âˆ´ Our assumption that 6

Sol:

= 13 [(7 Ã— 11) + 1]

= 13 [78]

i.e., 13 Ã— 78 cannot be a prime number because it has a factor 13.

âˆ´ 13 Ã— 78 is a composite number.

Also 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5

= 5 [7 Ã— 6 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 1]

which is also not a prime number. [âˆµ It has a factor 5]

Thus, â€˜7 Ã— 11 Ã— 13 + 13â€™ and â€˜7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 + 5â€™ are composite numbers.

Sol:

Time is taken by Ravi to arrive one round of the field = 12 minutes.

The L.C.M. of 18 and 12 gives the exact number of minutes after which they meet at the starting point again.

We have:

â‡’ 18 = 2 Ã— 3 Ã— 3 and 12 = 2 Ã— 2 Ã— 3

âˆ´ LCM of 18 and 12 = 2 Ã— 2 Ã— 3 Ã— 3 = 36

Thus, they will meet again at the starting point after 36 minutes.**Revisiting Irrational numbers**

A number is said to be an irrational number, if it cannot be expressed in the form p/q, where â€˜pâ€™ and â€˜qâ€™ are integers and q â‰ 0.

For example, , 0.10110111011110 ..., p, etc., are irrational.

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