I used to be able to do these

With 1,7,49
We get
J/K = (F-1/S)/(F-1/M) = 8/7
So K=7
T=KFM/2(M-F) = 49/12 = 4 1/12
After 1T , the planes have gone 4+1/12,1/2+1/12,1/12 so they are lined up FSXM
After 2T they have gone twice that, 8+1/6,1+1/6,1/6 so they are FMSX

@iamatiger

Here is a good one, the planets are lined up with periods 1, pi, and e. When do they next align?

@iamatiger said
@iamatiger

Here is a good one, the planets are lined up with periods 1, pi, and e. When do they next align?

Don't understand the question Tiger.
By 1,p1 and e do you mean 3 times ?

@venda

I mean the periods of the planets are 1, pi, and e, where pi and e are the fundamental constants. i.e the periods of the planets are 1 year, about 2.718 years, and about 3.14 years. When will they next all be in line with the sun?

@iamatiger said
@iamatiger

Here is a good one, the planets are lined up with periods 1, pi, and e. When do they next align?

Since J / K must be rational they will never align

@forkedknight
Correct! This is why there is no simple equation that solves the problem with 3 planets.

@forkedknight said
Since J / K must be rational they will never align

Well spotted .
Pi and Eulers constant
I'll try to get a life instead of trying to find simple equations where non exist.
I've enjoyed these discussions and hope I've learned something!!
All the other problems in my old book are far too easy because even I can do them!!

@venda said
Well spotted .
Pi and Eulers constant
I'll try to get a life instead of trying to find simple equations where non exist.
I've enjoyed these discussions and hope I've learned something!!
All the other problems in my old book are far too easy because even I can do them!!

Sometimes in math you just need to know the bridge.

I am guessing these formulas assume speed is constant.

@eladar said
Sometimes in math you just need to know the bridge.

I am guessing these formulas assume speed is constant.

they assume perfectly circular, co-planar orbits

@forkedknight said
they assume perfectly circular, co-planar orbits

Ah so not actual orbits at all. I was thinking elliptical orbits with changing velocities.

@eladar said
Ah so not actual orbits at all. I was thinking elliptical orbits with changing velocities.

I think actual orbits would be far too complex for the man in the street(i.e us)One look at the site mentioned by forkedknight(SYZYGY) convinced me

@venda

If the last problem was based on a circle, seems to me the arc tangents being equal would be a nice generalization.

Here is a nice puzzle for those who do not know how to approach it. If Keven can do a job in 5 hours and Larry can do the same job in 3 hours, how long will it take them to do the job working together?

@eladar said
Here is a nice puzzle for those who do not know how to approach it. If Keven can do a job in 5 hours and Larry can do the same job in 3 hours, how long will it take them to do the job working together?

That depends heavily on how much the job can be parallelized. Are you assuming 100%?

@forkedknight said
That depends heavily on how much the job can be parallelized. Are you assuming 100%?

Yes,no loss of efficiency.

@eladar said
Here is a nice puzzle for those who do not know how to approach it. If Keven can do a job in 5 hours and Larry can do the same job in 3 hours, how long will it take them to do the job working together?

Approximately 1 hour and 53 minutes. Keven does 0.333 percent of the job each minute and Larry does 0.555 percent each minute. Together they complete 0.888 per cent per minute. The total job is finished after 112.6 minutes, or 1 hour and 53 minutes (minus a few seconds).