If p and q are two distinct prime numbers then their hcf is

  1. State whether the statement is True or False:The HCF of two distinct prime numbers is 1.
  2. 3) If p and q are two distinct prime numbers, then their HCF is


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State whether the statement is True or False:The HCF of two distinct prime numbers is 1.

The correct option is A True Finding the HCF of two distinct prime numbers: Prime Numbers:The numbers which have only two factors 1 and the number itself are called Prime Numbers. For example: 2 , 3 , 5 , 7 , 11 , 13 etc. Lets take any two prime numbers say 2 and 3, On factorising 2 and 3, using prime factorisation method,we get, The prime factorization of 2 = 2 × 1 The prime factorization of 3 = 1 × 3 The common prime factor is 1 Therefore, the required H . C . F . = 1 Lets take any other two prime numbers say 5 and 11, On factorising 5 and 11, using prime factorisation method,we get, The prime factorization of 5 = 5 × 1 The prime factorization of 11 = 1 × 11 The common prime factor is 1 Therefore, the required H . C . F . = 1 Hence, the HCF of two distinct prime numbers is always 1. Hence, the given statement is True.

3) If p and q are two distinct prime numbers, then their HCF is

Views: 5,958 3 m or 3 m + 1 for some integer m. [Hint : Let x be any positive integer then it is of the form 3 q , 3 q + 1 or 3 q + 2. Now square each of these and show that they can be rewritten in the form 3 m or 3 m + 1.] 5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9 m , 9 m + 1 or 9 m + 8 Views: 5,125 4 n = ( 2 ) 2 n; so the only prime in the factorisation of 4 n is 2 . So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4 n. So, there is no natural number n for which 4 n ends with the digit zero. You have already learnt how to find the HCF and LCM of two positive integers the Fundamental Theorem of Arithmetic in earlier classes, without realising it! through an example. Example 6: Find the LCM and HCF of 6 and 20 by the prime factorisation method. Solution : We have : 6 = 2 1 × 3 1 and 20 = 2 × 2 × 5 = 2 2 × 5 1. Views: 5,686 c o t 2 5 ∘ t a n 6 5 ∘ ​ . Solution : We know : So, i.e., c o t 2 5 ∘ t a n 6 5 ∘ ​ = t a n 6 5 ∘ t a n 6 5 ∘ ​ = 1 Example 10: If sin 3 A = cos ( A − 2 6 ∘ ), where 3 A is an acute angle, find the of A. Solution : We are given that sin 3 A = cos ( A − 2 6 ∘ ). Since sin 3 A = cos ( 9 0 ∘ − 3 A ), we can write ( 1 ) as cos ( 9 0 ∘ − 3 A ) = cos ( A − 2 6 ∘ ) Since 9 0 ∘ − 3 A and A − 2 6 ∘ are both acute angles, therefore. 9 0 ∘ − 3 A A ​ = A − 2 6 ∘ = 2 9 ∘ ​

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