# Need some math home work help
**URL:** https://boards.straightdope.com/t/need-some-math-home-work-help/97140
**Category:** Factual Questions
**Created:** [December 6, 2001, 1:00am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140 "2001-12-06T01:00:23Z")
**Posts on this page:** 9
**Page:** 1
### Author:  [@astro](https://boards.straightdope.com/u/astro)
#### Post date: [December 6, 2001, 1:00am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/1 "2001-12-06T01:00:23Z")
How humiliating. I’m so tired.
I’m trying to help my 6th grader with a math problem. I got 3 hours of sleep last night and I can’t really remember or focus on the best way to do this regarding the most efficient way to calculate least common multiples. I could hammer away with the calculator for the next hour to try to find a common divisor but I know there has got to be a better way to do this.
Here is the problem:
“Suppose Earth and Mars are aligned with the sun. Earth completes it’s orbit in 365 days and Mars completes its orbit in 687 day (orbits rounded to the nearest Earth day).
When do both planets return to these same positions in their orbits?”
Any help re most efficient prime factorization method appreciated.
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### Author:  [@Manlob](https://boards.straightdope.com/u/Manlob)
#### Post date: [December 6, 2001, 1:19am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/2 "2001-12-06T01:19:00Z")
365 is easily found to factor into 5_73. 687 is not a multiple of 5 or 73, so least common multiple is 5_73\*687.
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### Author:  [@ModernRonin2](https://boards.straightdope.com/u/ModernRonin2)
#### Post date: [December 6, 2001, 1:20am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/3 "2001-12-06T01:20:22Z")
[http://www.google.com](http://www.google.com) is your friend:
[http://www.bmcc.org/nish/MathTutorials/Numbers/n-lcm.htm](http://www.bmcc.org/nish/MathTutorials/Numbers/n-lcm.htm)
Basically, factor the numbers down to primes. Then take a certain combination (too complicated to explain in a quickie post, but really not very hard) and the answer drops right out.
```
-Ben
```
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### Author:  [@Cabbage](https://boards.straightdope.com/u/Cabbage)
#### Post date: [December 6, 2001, 1:23am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/4 "2001-12-06T01:23:25Z")
Do you remember how to figure out the least common multiple if the prime factorizations are given? That’s one way to do it, but not what I would recommend.
Probably the most efficient method is the Euclidean algorithm; it’s probably easier to demonstrate than explain. If you don’t catch the pattern, just ask:
687 / 365 = 1, remainder 322
365 / 322 = 1, remainder 43
322 / 43 = 7, remainder 21
43 / 21 = 1, remainder 1
21 / 1 = 21, remainder 0.
The last nonzero remainder is the greatest common divisor, which is 1 in this case.
To get the least common multiple, multiply the two numbers together, then divide by their greatest common divisor:
(365 \* 687) / 1 = 250755.
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### Author:  [@astro](https://boards.straightdope.com/u/astro)
#### Post date: [December 6, 2001, 1:42am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/5 "2001-12-06T01:42:46Z")
Thanks to everyone. I had calculated the 229x3 and 73x5 factors but had figured there _had_ to be some intermediate factor common to both smaller than simply multiplying the numbers together. Apparently there is not so the answer is (I think) 687 years as to when they return to precisely the same solar alignment, unless I’m mis-reading the question (entirely possible given my current mental state).
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [December 6, 2001, 4:31am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/6 "2001-12-06T04:31:53Z")
FWIW, here’s the Euclidean algorithm for finding the GCD of two integers _a_ and _b_. Let _a_ be the greater number.[list=1][li] If _b_ is equal to zero, the GCD is _a_. Finish.[/li] Otherwise, change _b_ to _a_ mod _b_, and change _a_ to _b_. Then go back to step 1.[/list=1]Of course, _a_ mod _b_ is the remainder upon dividing _a_ by _b_.
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [December 6, 2001, 8:01am UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/7 "2001-12-06T08:01:42Z")
Of course, the numbers were rounded off to make the problem work out. The Earth actually takes 365.2563604 days to orbit the sun.
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### Author:  [@SteverinoAlaReno](https://boards.straightdope.com/u/SteverinoAlaReno)
#### Post date: [December 6, 2001, 4:20pm UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/8 "2001-12-06T16:20:01Z")
I’m a little confused. The earth would make one full revolution while Mars made little more than half a revolution. Eventually, the earth would “catch up” with Mars before Mars completed its first revolution, right? And then the two planets would again be aligned with the sun. That would take a lot less than 687 years. This problem, in fact, would be like calculating when the minute hand on a clock is aligned with the hour hand.
Now, if the question is intended to read when the planets would be in the exact same position relative to the sun, that’s a different story, and the previous answers would be correct.
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### Author:  [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)
#### Post date: [December 6, 2001, 4:36pm UTC](https://boards.straightdope.com/t/need-some-math-home-work-help/97140/9 "2001-12-06T16:36:11Z")
> [@](#):
>
> \*Originally posted by SteverinoAlaReno \*
> \*\*Now, if the question is intended to read when the planets would be in the exact same position relative to the sun, that’s a different story, and the previous answers would be correct. \*\*
The OP does say “same positions in their orbits” so I’d go with the previous answers. Still, yours is a more interesting question. In fact, I was looking at it last year, [for **manhattan’s** favorite question of the week](http://boards.straightdope.com/sdmb/showthread.php?threadid=39487). It’s the same as calculating how often Mars gets close to the Earth. And when the perihelion of Mars coincides with it’s closest apporach to the Earth, it appears very bright to us.