Planets A, B and C orbit a certain star once every 3, 7, and 18 months respectively. If the three planets are now in the same straight line, what is the smallest number of months that must pass before they line up again?
The least common multiple is 126.
thank you but it is not on my answer choice
What are your answer choices?
the answer choises are a)20 1/5 b)21 c)20 5/25 d)22
plese help i am good at math but i really don't get this i would appreciate if you can help me
r u there
To find the smallest number of months that must pass before the three planets line up again, we need to determine the least common multiple (LCM) of the orbital periods of the three planets.
The LCM is the smallest positive integer that is divisible by all the given numbers. In this case, the orbital periods are 3, 7, and 18 months.
To find the LCM, we can follow these steps:
1. List the prime factors of each number:
- 3 = 3^1
- 7 = 7^1
- 18 = 2^1 × 3^2
2. Take the highest power of each prime factor that occurs in any of the numbers:
- 2^1 × 3^2 × 7^1
3. Multiply all the prime factors together to get the LCM:
LCM = 2^1 × 3^2 × 7^1
Simplifying the product, we get:
LCM = 2 × 3^2 × 7 = 2 × 9 × 7 = 126
Therefore, the smallest number of months that must pass before the three planets line up again is 126 months.