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Class 10MathematicsReal NumbersHCF and LCM using prime factorisationHard
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A number leaves remainder (4) when divided by (21), (28), and (36). Which is the smallest such number?

Answer and explanation

Correct answer: (256)

Step 1: After subtracting (4), the number must be divisible by all three numbers. Step 2: The LCM of (21), (28), and (36) is (252), so the number is (252+4=256). Step 3: In common-remainder questions, find the LCM first.

Related tags

Real-NumbersLcmRemainderHcf And Lcm Using Prime FactorisationHcf Lcm Using Prime FactorisationReal NumbersMathematicsClass 10 Mcq
सुधार सुझाएँ / Earn

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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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