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?
hint: what is the LCM of 3,7, and 18 ?
72
72/7 = ?
Try again.
Oops I was thinking of 8 times 9
It's 126
Right.
To determine the smallest number of months that must pass before Planets A, B, and C line up again, we need to find the least common multiple (LCM) of their orbital periods.
The orbital periods of Planets A, B, and C are 3, 7, and 18 months, respectively.
First, let's find the LCM of 3 and 7:
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30...
Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70...
The least common multiple for 3 and 7 is 21 months.
Next, let's find the LCM of 21 and 18:
Multiples of 21: 21, 42, 63, 84, 105, 126, 147, 168, 189, 210...
Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180...
The least common multiple for 21 and 18 is 126 months.
Therefore, the smallest number of months that must pass before Planets A, B, and C line up again is 126 months.
To calculate this, you could also use the formula for finding the LCM of two numbers, and repeat the process for all three numbers:
LCM(a, b, c) = LCM(LCM(a, b), c)
LCM(LCM(3, 7), 18) = LCM(21, 18) = 126