写成质因数分解的形式,p=33×5×72p = 3^3 \times 5 \times 7^2。

  1. (a)

    将 450 表示为质因数分解的形式。

    答:___

  2. (b)

    求 pp 和 450 的最大公因数。

    答:___

  3. (b)

    求最小的整数 mm,使得 450mm 是 pp 的倍数。

    答:m=m = ___

  4. (c)

    求 aa 和 bb 的值,使得 p×abp \times \frac{a}{b} 是一个完全立方数,其中 aa 和 bb 均为质数。

    答:a = ___

    b = ___

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完整解法

4(a)

1. Break 450 down into prime factors.

450=45×10450 = 45 \times 10

=(3×3×5)×(2×5)= (3 \times 3 \times 5) \times (2 \times 5)

2. Write the prime factors in index form.

450=2×3×3×5×5450 = 2 \times 3 \times 3 \times 5 \times 5

=2×32×52= 2 \times 3^2 \times 5^2

答案

450=2×32×52450 = 2 \times 3^2 \times 5^2

4(b)

1. Compare the prime factorisations of $p$ and 450 and take the common factors with the lowest powers.

p=33×5×72p = 3^3 \times 5 \times 7^2

From part (a),
450=2×32×52450 = 2 \times 3^2 \times 5^2

Common prime factors: 33 and 55.

So the HCF is
32×5=45.3^2 \times 5 = 45.

答案

225225

4(b)

1. Compare the prime powers in $450$ with those needed for $p$.

Given
p=33×5×72p = 3^3 \times 5 \times 7^2
and from part (a),
450=2×32×52.450 = 2 \times 3^2 \times 5^2.
For 450m450m to be a multiple of pp, it must contain at least 33×5×723^3 \times 5 \times 7^2.

450450 already has 323^2 and 525^2, so it still needs one more factor of 33 and two factors of 77.

2. Find the smallest possible value of $m$.

So the smallest integer mm is
m=3×72=147.m = 3 \times 7^2 = 147.

答案

m=147m = 147

4(c)

1. Identify how the prime powers in $p$ must change to become a perfect cube.

Given
p=33×5×72p=3^3\times 5\times 7^2.

For a perfect cube, every prime factor must have an exponent that is a multiple of 33.

333^3 is already suitable.

515^1 should be removed by dividing by 55.

727^2 should be multiplied by 77 to become 737^3.

2. Choose the prime numbers $a$ and $b$ and verify the product is a perfect cube.

Hence
ab=75.\frac{a}{b}=\frac{7}{5}.

Then
p×75=33×5×72×75=33×73,p\times \frac{7}{5}=3^3\times 5\times 7^2\times \frac{7}{5}=3^3\times 7^3,
which is a perfect cube.

So a=7a=7 and b=5b=5.

答案

a=7, b=5a=7,\ b=5

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