Problem 1:

Inez waters her plants every two days. She trims them every 15 days. She did both today. When will she do both again?

Problem 2:

An airport offers two shuttles that run on different schedules. If both shuttles leave the airport at 4:00 P.M., at what time will they next leave the airport together?

[tex]\[
\begin{tabular}{|c|c|}
\hline
\multicolumn{2}{|c|}{\textbf{Shuttle Schedule}} \\
\hline
\textbf{Shuttle} & \textbf{Departs} \\
\hline
A & every 6 minutes \\
\hline
B & every 9 minutes \\
\hline
\end{tabular}
\][/tex]

Hint: To find the next time both shuttles leave together, calculate the least common multiple (LCM) of their departure intervals.



Answer :

Let's break this down step by step.

### Problem 1: When will Inez water and trim her plants again on the same day?

1. Identify the intervals:
- Inez waters her plants every 2 days.
- Inez trims her plants every 15 days.

2. Find the Least Common Multiple (LCM):
- To determine when both events coincide (watering and trimming on the same day), we need to find the LCM of 2 and 15.

3. Calculate the LCM of 2 and 15:
- The prime factors of 2 are [tex]\(2\)[/tex].
- The prime factors of 15 are [tex]\(3 \times 5\)[/tex].

4. Combine the highest powers of these prime factors:
- The LCM is the product of the highest powers of all prime numbers appearing in the factorization, which gives us: [tex]\(2 \times 3 \times 5 = 30\)[/tex].

Therefore, Inez will water and trim her plants again on the same day after 30 days.

### Problem 2: At what time will the two shuttles leave together again?

1. Identify the intervals:
- Shuttle A departs every 6 minutes.
- Shuttle B departs every 9 minutes.

2. Find the LCM:
- To determine when both shuttles will depart at the same time again, we need to find the LCM of 6 and 9.

3. Calculate the LCM of 6 and 9:
- The prime factors of 6 are [tex]\(2 \times 3\)[/tex].
- The prime factors of 9 are [tex]\(3^2\)[/tex].

4. Combine the highest powers of these prime factors:
- The LCM is the product of the highest powers of all prime numbers appearing in the factorization, which gives us: [tex]\(2 \times 3^2 = 18\)[/tex].

Therefore, the two shuttles will leave together again after 18 minutes.

Given both shuttles leave the airport at 4:00 P.M., they will next leave together at:

[tex]\[ 4:00 \text{ P.M.} + 18 \text{ minutes} = 4:18 \text{ P.M.} \][/tex]

So, the shuttles will next leave the airport together at 4:18 P.M.