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36, 45 and 54 are three given numbers. i) Find the H.C.F. of these numbers. ii) Find the greatest number that can be added to 2 so that the sum divides these numbers exactly.

Mathematics

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2 parts
i

9\text{9}

ii

7\text{7}

Step-by-step solution

7 steps

7. 36, 45 and 54 are three given numbers.

Part i· 3 steps

Find the H.C.F. of these numbers.

  1. Find the prime factorization of each number.

    36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 45=3×3×5=32×5145 = 3 \times 3 \times 5 = 3^2 \times 5^1 54=2×3×3×3=21×3354 = 2 \times 3 \times 3 \times 3 = 2^1 \times 3^3

  2. Identify the common prime factors and their lowest powers.

    The common prime factor is 3. The lowest power of 3 present in all factorizations is 323^2.

  3. Multiply the common prime factors with their lowest powers to find the H.C.F.

    H.C.F.=32=3×3=9\text{H.C.F.} = 3^2 = 3 \times 3 = 9

    The H.C.F. of 36, 45, and 54 is 9\boxed{\text{9}}.

Part ii

7\text{7}

Find the greatest number that can be added to 2 so that the sum divides these numbers exactly.

  1. Understand the problem.

    We are looking for a number, let's call it XX. When XX is added to 2, the sum (X+2)(X+2) must divide 36, 45, and 54 exactly. This means (X+2)(X+2) is a common factor of these numbers.

  2. To find the greatest such number XX, the sum (X+2)(X+2) must be the greatest common factor (H.C.F.) of 36, 45, and 54.

  3. From part (i), we found that the H.C.F.

    of 36, 45, and 54 is 9.
    So, we set the sum equal to the H.C.F.:

    X+2=9X + 2 = 9

  4. Solve for XX.

    X=9−2X = 9 - 2 X=7X = 7

    The greatest number that can be added to 2 is 7\boxed{\text{7}}.

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