A Little Light Thinking


Zafirah and Folashade suggested some ideas for attacking the problem:
 
Do not use random numbers when solving this problem go up in a strategic method.  Before trying to work out anything at least get three numbers which light up each light to see if there is any pattern occouring. Look closely at the relationship between the numbers for each colour. Once you find out the relationship between the numbers, start to look if you are able to put the method into a formula.
 
 
Dylan and Ashara followed on with these thoughts:
 
Try to display the rules algebraicly. Our sequence which goes $3, 9, 15$ could be described as "going up by $6$ from the number $3$", or could be described as $6n - 3$, $n$ being the number in the sequence. If you're having difficulty displaying your results algebraically, why not try to imagine a numberline. $+ 1$ merely describes you shifting all the numbers up by one. Once you have found out the rules for Red & Blue, finding out the combined rule shouldn't prove difficult. At times, the combined rule could also be the rule for either Red or Blue! Just try loads of number, and don't think too hard. Sometimes the easiest, most simple answers are also the correct ones.  

 
Some of you found sequences which didn't ever light both lights. Josh, Christian, Pavan, Alex and Vishaal found one such scenario:
 
We concluded to the rules: $3n+1$ and $6n+2$. We have decided that the two are impossible to solve, as it will never both have the same number. The reason for that is because $3n+1$ will always be one more than a multiple of three whereas $6n+2$ will always two more then a multiple of three.
 
Rajeev of Fairfield Junior School gave some examples when both lights could be lit:
 
$8n+3$ and $6n+5$... the number which would turn both lights on is $11$ because $8+3$ and $6+5$ both equal $11$.
The other numbers which would work are $35, 83,131$.
The formula is $24n-13$.

$11n+10$ and $10n+1$... the number which would turn both lights on is $21$ because $11+10$ and $20+1$ both equal $21$.
The other numbers are $131,241,351$ and so on.
The formula is $110n-89$.

In the above you have to find the lowest common multiple of both sequences and that number minus another number so the resulting number would turn both lights on.

Jessica and Sam suggested some other combined sequences:

$4n+1$ and $5m-2$ have a combined rule of $20k-7$
$10n+4$ and $3m-1$ have a combine rule of $30k+14$
$11n-5$ and $6m-5$ have a combined rule of $66k-5$

A pattern we can see for the combined rules is that to find the combined rules times the gradients together and then find the common differences in the new sequence.   
 

Finally Herschel of the European School of Varese gave an insight into the method known as the Chinese Remainder Theorem:
 
We want to find a number that is part of two Arithmetic Progressions (APs): $an+b$ and $cn+d$. To put it another way, we are searching for a number $x$ that satisfies the congruences $x = b$ (mod$ a$) and $x = d$ (mod $c$). (The "mod $a$" part means that you look at the remainder on each side of the equation after dividing by "$a$".)

Finding this value of $x$ that will satisfy both equations and hence turn on both lights can be done using Chinese Remainder Theorem, which deals precisely with this problem - solving congruences.

For more information on the Chinese Remainder Theorem please look at this article.