Kids maths puzzle (factors, common multiples,...)???

esuhl

I'm trying to help my niece with her maths homework.  Here's the puzzle (which I've paraphrased to keep it short):

A planet has seven moons (A to G) on a single plane.  Moon A (the closest) takes one year to orbit the planet, Moon B takes two years, C takes 3, etc.  When all moons line up it is called a "super-eclipse".  How long is it between each super-eclipse?

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Here's my thinking so far:

The number of years between super-eclipses must be divisible by the numbers 1 to 7.  (Is that right?)

Seven factorial (7!, i.e. 7x6x5x4x3x2x1) is divisible by all the numbers from 1 to 7 because it is the multiple of all those numbers.

7! = 5040 years

But that's not the shortest time between eclipses.  I can divide 5040 by 2 to get 2520 years, which is still divisible by the numbers from 1 to 7.

I'm stuck on working out how to find the lowest number that is divisible by 1 to 7...

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I created a "brute force" simulation in Excel, which gives the answer of 420 years.  But I don't know how to calculate that without trial-and-error.

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A little googling suggests that it's something to do with finding the "least common multiple" and "prime factors"...  But it's leaving me feeling rather old and confused.  I'm not even sure calculating 7! is the right way to start...

Also... the method I'm thinking about using assumes that the planets only line up on complete cycles (i.e. in their starting positions).  Is it possible that they all line up at other points in their cycle?

Any maths experts out there?!  Cheers :-)

stragglebod

I would start by working out the factors of all the numbers from 1 to 7 and disregarding any factors that are also factors of the larger numbers. So for instance you can ignore 3 because its factors 1x3 are also factors of 6. Same for 1 (all of them), 2 (4,6).

Just done it on paper and noticed that leaves 4,5,6,7 but 4 and 6 have a common factor of 2 which you can ignore, so the calculation becomes 4x5x6x7/2

esuhl

Ah, typical!  I've been pondering this all day, but I think I've cracked it.

I need to find the "lowest common multiple" of the numbers 1 to 7, right.  And this web page explains how to do that:

https://www.calculatorsoup.com/calculators/math/lcm.php

I think I'm on the right track now! :-D

esuhl

stragglebod said:

I would start by working out the factors of all the numbers from 1 to 7 and disregarding any factors that are also factors of the larger numbers. So for instance you can ignore 3 because its factors 1x3 are also factors of 6. Same for 1 (all of them), 2 (4,6).

Just done it on paper and noticed that leaves 4,5,6,7 but 4 and 6 have a common factor of 2 which you can ignore, so the calculation becomes 4x5x6x7/2

Wow -- I had to re-read that a few times!  I get it now.  I couldn't have done it without you, but this is how I've got it in my head:

Find the factors of numbers 1 to 7:

1 = 1 x 1

2 = 2 x 1

3 = 3 x 1

4 = 4 x 1, 2 x 2

5 = 5 x 1

6 = 6 x 1, 3 x 2

7 = 7 x 1

Not relevant with such small numbers, but with bigger numbers, keep finding factors of factors, e.g.:

16 = 16x1, 8x2, 4x2x2, 2x2x2x2

Disregard all but the multiple with the smallest (therefore prime) numbers:

1 = 1 x 1

2 = 2 x 1

3 = 3 x 1

4 = 2 x 2

5 = 5 x 1

6 = 3 x 2

7 = 7 x 1

...

(16 = 2 x 2 x 2 x 2)

Then multiply each number from 1 to 7 IF it occurs to the right of an equals sign above:

1 x 2 x 3 x 5 x 7 = 210 years

Brilliant!  Thanks for your help!

esuhl

Oh... hang on... 210 isn't divisible by 4!  Whoops!

WaywardDriver

Express each number as multiples of prime numbers:
1 = 1; 2 = 1*2; 3 = 1*3; 4 = 1*2*2; 5 = 1*5; 6 = 1*2*3; 7 = 1*7
Then calculate the least common multiple of these prime numbers = 1*2*3*2*5*7 = 420 (with the 2nd 2 coming from 4=1*2*2)   
To check answer use Excel formula =LCM(1,2,3,4,5,6,7) to give 420.