Another ploy I have sometimes used to solve puzzles I am stuck with is to look at the answer(when given) and work out the formula(e) to get there.
In the meantime there is a formula that seems to work for the type of puzzles below for 2 planets but falls when a 3rd planet is introduced:-
2 planets orbit the sun in the same direction.The 1st planet takes 33 years and the second 9 years for a full orbit.They are now in a straight line with the sun.When will this next occur?
The formula is xy/2(x-y).
They do not have to be in the same position as at the start to be in a straight line.
I think that is the trap.
So you would think that for 3 planets you could calculate the answer for 2 and then just use the same formula to add the 3rd.
But it doesn't work .
Trying it for 6 years , 5 years and 2 years and the answer given in my old book is not what you get doing it this way.
I never quite got to the bottom of it
@wolfgang59said After 30 years they will be back where they started but the faster
planet will have done one more orbit than the slower. It therefore
will have been aligned once before. After 15 years.
Does it count if they're on opposite sides of their orbits? I assumed not.
I was only considering them aligned if there was a ray from the sun through both planets.
@wolfgang59said Yes it does.
If the question were when do they return to the same alignment it would
simply be the lowest common multiple of the two orbital periods.
I'm not convinced that formula is correct though!
Correct .If it was the lowest common multiple the problem would be easy.However,I think there were about 3 examples in the book and the formula worked for 2 planets for them all.
The planets do not have to be on either side of the sun which, as I said was the trap.
There must be a formula for "n" number of planets though.
I expect Eladar or Tiger will give us the answer eventually.