Thrice in a Blue Moon
On the planet Blue Ox, there are three moons: a red one, a blue one, and a regular one. The red moon orbits around the planet once every 66 days, the blue moon orbits around the planet every 3 days, and the regular moon orbits around the planet every 30 days. The inhabitants of Blue Ox love to throw pies at newcomers. They would usually throw their pies at the newcomer unless that the day is when their three moons line up in a row, because they think that day is unlucky. Josh the Astronaut has just arrived on Blue Ox. The probability that he would be hit with pies is a fraction b a . What is a+b?
The answer is 659.
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4 solutions
Your answer/solution is wrong. The moons do not necessarily line up only on days multiples of 330. For example, after 220 days, all three moons are lined up, they are all 1/3 of the way to finishing their orbit.
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Yes, there clearly is an issue with the wording of this problem. "Moons line up in a row" can be understood in many substantially different ways.
Good point there. Can u further explain your proof? Seems like almost everyone assumed the 3 moons meet at multiples of 330
Actually, there is an even faster way with just 110 days.
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Yup, 1 1 0 is the minimum number of days for the moons to line up in a row.
Yup, it's wrong. 1 1 0 is the minimum days required for the moons to line up in a row. And they do that every 1 1 0 days. So, the probability should be 1 1 0 1 0 9 . I did all the work for nothing :(
I'm thinking of requesting a dispute.
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Sent a dispute that reads like:
"The three moons actually line up in a row every 1 1 0 days [not 3 3 0 days as the answer suggests].
In 1 1 0 days, the red moon has an angular displacement of 6 6 3 6 0 × 1 1 0 = 6 0 0 ≡ 2 4 0 m o d 3 6 0 degrees.
In 1 1 0 days, the blue moon has an angular displacement of 3 3 6 0 × 1 1 0 = 1 3 2 0 0 ≡ 2 4 0 m o d 3 6 0 degrees.
In 1 1 0 days, the regular moon has an angular displacement of 3 0 3 6 0 × 1 1 0 = 1 3 2 0 ≡ 2 4 0 m o d 3 6 0 degrees.
That means every 1 1 0 days the moons will align. And our desired probability is 1 1 0 1 0 9 and a + b = 2 1 9 , not 6 5 9 as the answer suggests.
Thank you!"
Yes, you're right. If you call however many days have elapsed D , and the orbital period of a specific moon V i , then they are lined up whenever,
V 1 D m o d V 1 = V 2 D m o d V 2 = V 3 D m o d V 3
So they are not lined up only when D ≡ 0 ( m o d V i ) for all moons.
And this is also not taking into account the cases where the moons are lined up but on opposite sides of the planet.
This is funny. Sometimes, you have to be stupid to get the answer that another stupid person wants.
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Yes, thank you.
I hope you realize that I am 14 and that you are twice my age.
Does anyone have a pure maths solution to this (110 solution not 330) - perhaps using trigonometry? My own method was a bit clunky...
~clap clap~
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"Applause" is the shorter word
Just like there is a smaller answer...
good job
Count from the last day when the moons were aligned. For the number of cycles of each moon to be integral, the number of days since the last alignment must be divisible by the period of each of their cycles. Thus the closest alignment is at lcm ( 3 , 3 0 , 6 6 ) = 3 3 0 .
For 3 2 9 out of the 3 3 0 days between two alignments, Josh would get hit by pies. So the probability of him getting hit by pies is 3 3 0 3 2 9 ⇒ 3 2 9 + 3 3 9 = 6 5 9
You are wrong. Look at the comments against Ben Frankel to see why.
Good Job
~clap clap~
We first find the LCM of 66, 3, and 30, which is 330. Then, there is only one day in which they do not throw pies, so the other 329 days they do throw pies. So, the probability is 329/330, and then 329+330=659.
Bob, I'm sorry but your solution isn't right.
The moons actually line up every 1 1 0 days. In 1 1 0 days, the moons have an angular displacement of 2 4 0 m o d 3 6 0 degrees. The correct probability is 1 1 0 1 0 9 .
Try it yourself to see why you're wrong.
The reason is because your answer is incorrect. If the moons are on directly opposite side of the planet, they are still "lined up". Take this for example (pretend the words represent the object):
(redmoon) (bluemoon) (planet) (regularmoon).
The moons are still lined up, but this didn't necessarily happened on a multiple of 330.
View my post to Ben Frankel's solution to show more proof that your answer is wrong. Congratulations, Bob Yang, for posting a problem that had an incorrect solution.
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Congratulations, William Zhang, for using such mean sarcasm.
this shouldn't be too hard
I thought this would be a Level 2 question, but hmmmm...
why is its rating like 1991
now its rating is like 2077 wow
66=11 \times 3 \times 2;
3=3;
30=2 \times 5 \times 3;
Three moons line up every 330 days;
ans=659
The moons line up every 110 days. It's a good problem though Bob...
Imagine that all of the moons start aligned.
Now, 3 days later, the two slower moons have moved a bit but the fastest one has made an entire cycle. Then, 30 days after they had all started lined up, the fast moon and the medium speed moon are lined up back where they started, but the slow moon is still not back to the start.
The moons only line up after an amount of days passes that is a multiple of the speed of each planet. We will thus find the LCM of the three speeds:
lcm ( 3 , 3 0 , 6 6 ) = 2 ⋅ 3 ⋅ 5 ⋅ 1 1 = 3 3 0
So the moons all line up one out of 330 days. The probability that Josh the Astronaut will not get hit by pies is thus the probability that the moons won't be lined up on the day that he arrives, which is,
1 − 3 3 0 1 = 3 3 0 3 2 9 = b a
So the answer is,
a + b = 3 2 9 + 3 3 0 = 6 5 9