Plantets A, B, and C orbit a certain star once every 2, 5, and 12 months. If the three plantes are now in the same straight line, what is the smallest number of months that must pass before they line up again?
I do not want the answer, I just do not understand how to set up the equation.
yes
I think I figured it out...If I go through all the multiples of
2 5 and 12, the LCM would be 60, so wouldn't that be the number of months?
I believe the LCM solution is most probably the correct one.
However, if we interpret the expression "in the same straight line" in the general sense, it means that any planet could be on the same side OR on the other side of the sun. In which case the LCM would be for half the period, time it takes to go across the other side of the sun.
If the general interpretation is accepted, we can say that the planets will be on a straight line again in LCM(1, 2.5, 6)=15 months.
change "sun" for "star".
Oh, lining up planets, I see! Well, let's break this down. To find the smallest number of months that must pass before the planets line up again, we need to find the least common multiple (LCM) of their orbital periods.
So, let's start by finding the LCM of 2, 5, and 12. One way to do this is to prime factorize each number and multiply the highest power of each prime factor.
Let's start with 2: 2 = 2^1
Next is 5: 5 = 5^1
Finally, 12: 12 = 2^2 * 3^1
Now, we take the highest power of each prime factor and multiply them together:
highest power of 2 = 2^2
highest power of 3 = 3^1
highest power of 5 = 5^1
Multiply them together: 2^2 * 3^1 * 5^1 = 60
So, the smallest number of months that must pass before the planets line up again is 60 months.
To figure out when the three planets will line up again, we need to find the least common multiple (LCM) of their orbital periods. The LCM is the smallest number that is divisible by all the given numbers.
In this case, the orbital periods of planets A, B, and C are 2 months, 5 months, and 12 months, respectively. To find the LCM, we can follow these steps:
1. First, write down the prime factorization of each orbital period:
- For 2 months, the prime factorization is 2^1.
- For 5 months, the prime factorization is 5^1.
- For 12 months, the prime factorization is 2^2 * 3^1.
2. Write down the highest power of each prime factor:
- The highest power of 2 is 2^2.
- The highest power of 3 is 3^1.
- The highest power of 5 is 5^1.
3. Multiply the highest powers:
2^2 * 3^1 * 5^1 = 4 * 3 * 5 = 60.
Therefore, the smallest number of months that must pass before the three planets line up again is 60 months.