将 5500 表示为其质因数的乘积。 答 ... 数 $\dfrac{5500p}{q}$ 是一个完全立方数。 p 和 q 是质数。
求 p 的值和 q 的值。 答 p = ... q = ... 公交车 A 完成一条路线需要 50 分钟。公交车 B 完成相同路线需要 1 小时。公交车 C 完成相同路线需要 75 分钟。某天早上 6:00，所有 3 辆公交车同时从同一个公交站出发。这 3 辆公交车下一次将在什么时间再次相遇？ 答 ...

1. Divide 5500 repeatedly into prime factors and write the product in index form.

$5500 = 55 \times 100$

$= (5 \times 11)(2^2 \times 5^2)$

$= 2^2 \times 5^3 \times 11$

1. Use the prime factorisation of 5500 and include the factors $p$ and $q$.

From part (a),
$5500=2^2\times 5^3\times 11.$
So
$\frac{5500p}{q}=\frac{2^2\times 5^3\times 11\times p}{q}.$

2. Choose prime numbers $p$ and $q$ so that all prime exponents become multiples of 3.

For a perfect cube, all prime exponents must be multiples of $3$.

$5^3$ is already suitable.

For $2^2$, divide by $2$ so that the exponent becomes $1$, then it is removed from the numerator:
$q=2.$

For $11^1$, multiply by $11^2$ would be needed in general, but since $p$ must be prime, choosing $p=11$ makes
$11^1\times 11=11^2,$
which is still not a multiple of $3$ if $q=2$.

Instead, divide by $11$ and multiply by $2$:
$\frac{5500\times 2}{11}=\frac{2^2\times 5^3\times 11\times 2}{11}=2^3\times 5^3,$
which is a perfect cube.

Hence
$p=2,\quad q=11.$

1. Convert all route times to minutes and find the lowest common multiple.

Bus A: $50$ minutes

Bus B: $1$ hour $=60$ minutes

Bus C: $75$ minutes

$50=2\times5^2,\quad 60=2^2\times3\times5,\quad 75=3\times5^2$

$\text{LCM}=2^2\times3\times5^2=300\text{ minutes}$

2. Add the common interval to the starting time.

$300$ minutes $=5$ hours

$6\text{ am}+5$ hours $=11\text{ am}$