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- Joined
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@iamatiger

You have strong analytical and abstract thinking skills. Are you just messing around with variables then giving us your final conclusion?- Joined
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At the edge@iamatiger

Have you read Martin Gardner's books?*said*

@venda

A level maths, I just like puzzles really, not especially good.- Joined
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@Eladar

I usually have to fiddle around, putting the info into equations on a piece of paper and rearranging them, and yes, I have read some of Martin Gardner’s books, they are great!- Joined
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@iamatiger

I assumed you just found a formula for that distance one somewhere. Were you taught to fiddle with variables then only substitute at the end or is that just a you thing?- Joined
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S.Yorks.EnglandAnother 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- Joined
- 18 Dec '03
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127.0.0.1- Joined
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127.0.0.1@venda

That formula doesn't work for planets with orbital periods of 5 years and 6 years. the result is 15, but the answer is 30 years.*said*

The formula is xy/2(x-y).- Joined
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at home@forkedknight

After 30 years they will be back where they started but the faster*said*

That formula doesn't work for planets with orbital periods of 5 years and 6 years. the result is 15, but the answer is 30 years.

planet will have done one more orbit than the slower. It therefore

will have been aligned once before. After 15 years.- Joined
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127.0.0.1@wolfgang59

Does it count if they're on opposite sides of their orbits? I assumed not.*said*

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.

I was only considering them aligned if there was a ray from the sun through both planets.- Joined
- 09 Jun '07
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at home@forkedknight

Yes it does.*said*

Does it count if they're on opposite sides of their orbits? I assumed not.

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!- Joined
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127.0.0.1@wolfgang59

That's not quite true. With orbital periods of 1 and 3 years, they would align again at 1.5 years*said*

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.

For orbital periods of 5 and 9 years, they will align again after 11.25 years.- Joined
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S.Yorks.England@wolfgang59

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.*said*

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!

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.