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Class 10MathematicsReal NumbersHCF and LCM using prime factorisationExpertLevel 10
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The HCF of two numbers is (18) and their LCM is (1260). How many unordered pairs of such numbers are possible?

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Answer and explanation

Correct answer: (4)

Step 1: Let the numbers be (18m) and (18n), where (m) and (n) are coprime. Step 2: (18mn=1260), so (mn=70=2\times5\times7); this gives (4) unordered coprime factor pairs. Step 3: For a square-free product, split prime factors into two groups to count unordered pairs.

Related tags

Real-NumbersHcf-LcmPair-Count

Frequently asked questions

What is the correct answer to this question?

(4)

Why is this the correct answer?

Step 1: Let the numbers be (18m) and (18n), where (m) and (n) are coprime. Step 2: (18mn=1260), so (mn=70=2\times5\times7); this gives (4) unordered coprime factor pairs. Step 3: For a square-free product, split prime factors into two groups to count unordered pairs.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: HCF and LCM using prime factorisation.

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10 more MCQs
  1. 1If the product of two numbers is 3780 and their LCM is 315, what is their HCF?
  2. 2What is the HCF of (18) and (24) using prime factorisation?
  3. 3What is the LCM of (18) and (24) using prime factorisation?
  4. 4What is the HCF of (40) and (60)?
  5. 5What is the LCM of (40) and (60)?
  6. 6What will be the HCF of (45) and (75)?
  7. 7What will be the LCM of (45) and (75)?
  8. 8Which is the HCF of (32) and (48)?
  9. 9Which is the LCM of (32) and (48)?
  10. 10What is the HCF of (27) and (36)?
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