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Find the HCF of 43, 91 and 183. 2. What is the LCM of 24, 36 and 40? 3. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one number

Mathematics

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1. Find the HCF of 43, 91 and 183.

  1. Find the prime factorization of each number.

    43=43(43 is a prime number)43 = 43 \quad (\text{43 is a prime number}) 91=7×1391 = 7 \times 13 183=3×61183 = 3 \times 61

  2. Identify common prime factors.

    The numbers 43, 91, and 183 have no common prime factors other than 1.

  3. The HCF is the product of the common prime factors.

    Since there are no common prime factors other than 1, the HCF is 1.

    HCF(43,91,183)=1\text{HCF}(43, 91, 183) = 1

    The HCF of 43, 91, and 183 is 1\boxed{1}.

    2. What is the LCM of 24, 36 and 40?

  4. Find the prime factorization of each number.

    24=2×2×2×3=23×3124 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1 36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 40=2×2×2×5=23×5140 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1

  5. For each prime factor, take the highest power that appears in any of the factorizations.

    • For prime factor 2: The highest power is 232^3 (from 24 and 40).
    • For prime factor 3: The highest power is 323^2 (from 36).
    • For prime factor 5: The highest power is 515^1 (from 40).

  6. Multiply these highest powers to find the LCM.

    LCM(24,36,40)=23×32×51\text{LCM}(24, 36, 40) = 2^3 \times 3^2 \times 5^1 LCM(24,36,40)=8×9×5\text{LCM}(24, 36, 40) = 8 \times 9 \times 5 LCM(24,36,40)=72×5\text{LCM}(24, 36, 40) = 72 \times 5 LCM(24,36,40)=360\text{LCM}(24, 36, 40) = 360

    The LCM of 24, 36, and 40 is 360\boxed{360}.

    3. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one number is 275, then find the other number.

  7. Use the relationship between HCF, LCM, and the product of two numbers.

    For any two positive integers aa and bb:

    a×b=HCF(a,b)×LCM(a,b)a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)

  8. Substitute the given values into the formula.

    Let the first number be a=275a = 275.
    Let the second number be bb.
    Given HCF = 11.
    Given LCM = 7700.

    275×b=11×7700275 \times b = 11 \times 7700

  9. Solve for bb.

    275×b=84700275 \times b = 84700 b=84700275b = \frac{84700}{275} b=308b = 308

    The other number is 308\boxed{308}.

    4. List 5 pairs of relative numbers.

    Relative numbers (also known as coprime or relatively prime numbers) are two integers whose greatest common divisor (HCF) is 1.

    Here are 5 pairs of relative numbers:

    • (2,3)(2, 3)
    • (4,5)(4, 5)
    • (7,10)(7, 10)
    • (9,16)(9, 16)
    • (15,22)(15, 22)
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