Problem Teachers' Notes Hint Solution

Red Light:        the numbers go up in $5$'s starting at three. e.g $3,8,13,18...$

Yellow Light:   the odd numbers. e.g $1,3,5,7,9,11...$

Green Light:     the $11$ times table. e.g $11,22,33,44,55...$

Blue Light:       the $4$ times table. e.g $4,8,12,16,20 ...$

 

It is impossible to get them all to light up because in the 4 times table there is no odd number so you can never light up the Yellow light and the Blue light together.

 

 

RED          =  $3n$

YELLOW =  $3n-1$

GREEN    =  $9n-7$

BLUE       =  $11n-8$

 

So...RED could be any number, so lets call it $x$. And...YELLOW would be one less always, so $x-1$. This means that these numbers would always be 1 different and so all four numbers could never be lit up at the same time!!! 

 

 

 

Red     =  $2N-1$

Yellow =  $7N- 3$

Green   =  $10N+1$

Blue     =  $2N-1$

 

It is possible to get all colours light because and we can do it with the number $11$. It is $1$ less than numbers in the $2$ times table. It is $3$ less than numbers in the $7$ times table. It is $1$ more than numbers in the $10$ times table.

 

 

RULES: Testing whether the numbers are part of certain arithmetic progressions (APs).

RED:          $x=5+11n$  i.e. $x=5(mod11)$

YELLOW: $x=9n$        i.e. $x=0(mod9)$

GREEN:    $x=3+8n$    i.e. $x=3(mod8)$

BLUE:       $x=1+2n$    i.e. $x=1(mod2)$ (i.e. x is odd)

 

Solving by first principles (I actually used Chinese Remainder Theorem, but this is simpler!): This problem can first be simplified.

 

The blue light will light up if the number is odd, but as all numbers in the green AP are odd, we can eliminate the blue AP. Also, the yellow AP is simply the multiples of $9$. Therefore, we need to find odd multiples of $9$ that are in the red and green APs.

 

By trial and error, we "luckily" find that $9\times3=27$ is in all four APs, and is one of many answers.

 

There are other solutions, which occur every $792$, so the general solution would be $x=27+792n$. This is because $792$ a multiple (the LCM) of $11, 9, 8$ and $2$, so adding it to an existing solution will always give a new one. Thus, after $27$, the next solution would be $819$.