1. Pathways
  2. Exploring Our Number System
  3. Remainders
  4. Charlie's Delightful Machine

Charlie's Delightful Machine

Circuit board


Charlie's Delightful Machine has four coloured lights on it.
You can enter numbers into Charlie's machine.
Each light is controlled by a rule, and if your number satisfies the rule, the light will go on.
Some numbers may turn on more than one light!

Type in some numbers and see which lights you can switch on.
To start again with a new set of rules, click a Level button (each level is more difficult than the last).




Can you develop a strategy to work out the rules controlling each light at Level 1?
Once you have a strategy, challenge yourself to find four digit numbers that turn on each light.

Once you are confident that you can work out the rules for Level 1 lights, try finding rules for some Level 2 and 3 lights.

You may also want to try A Little Light Thinking.

 

 

Comments

Submitted by Vicky and Abby on Tue, 06/07/2016 - 07:01

Blue: 6n-4
Red: 11n-2
Yellow: 11n-4
Green:12n-2

Submitted by cfg21 on Wed, 06/08/2016 - 19:14

Do you think it would be possible to find a number that would light up all four lights at the same time?
If not all four, what about three lights? Or two lights?

Can you explain why some pairs of lights will never light up at the same time?

Submitted by Katy Thomas on Fri, 06/24/2016 - 08:42

I'm not sure, but I did try to investigate. You can light up two lights at a time, but not three, and the Green and Yellow lights never lit up together in my problem because one was even numbers and one was odd. You could try looking for patterns like that.

In some cases, there is probably a number in millions (or minus millions) that lights up all the numbers but, unless you are willing to sit at a computer screen for the rest of your life, you'll probably never know!

Submitted by Chung Dik Him, ... on Tue, 06/07/2016 - 07:19

Yellow $= 2n^2 - 2n + 1$

Blue $= 2n^2 + 4n - 4 $

Red $= 2n^2 - 4n + 3$

Green $= \frac{1}{2}n^2 - \frac{1}{2}n + 10$

Four Digit Numbers:

Yellow = 7081

Blue = 7436

Red = 6963

Green = 1780

Submitted by RM on Sun, 06/12/2016 - 13:32

For the blue light we have
5, 7, 9, 15, 19, 29, 35, 49 ,57
The only recurrence relation we have found is S(n+3) = 2( S(n+2) - S(n+1) ) + S(n) which only works for n = 2, 4, 6 !
Please help

Submitted by cfg21 on Fri, 06/24/2016 - 12:35

Can you find a rule for your first, third, fifth, seventh... values: 5, 9, 19, 35, 57?

What happens if you substitute negative numbers into your rule?

Does it help to draw a graph of your quadratic to see what is going on?

Submitted by Rachael Whitney... on Mon, 06/13/2016 - 04:33

I generated the following results (Level 1):

Yellow: 6, 16, 26,.... 10n+6
Red: 2, 10, 18, 26, 34,.... 8n+2
Blue: 3, 7, 11, 15, 19, 23, 27, 31.... 4n + 3
Green: 5, 14, 23, 32,.... 9n + 5

So to light up all colours the number must be an element of each of the above sets, so can a single number be an element of each of the sets above?

Red and Blue lights will not light up together as for any n:
4n + 3 is a multiple of 4 + 3, and 8n+ 2 = 4(2n) + 2 is a multiple of 4 + 2
and we can't have a number that is even and odd at the same time.

Love this challenge.

Submitted by Eliot on Mon, 06/13/2016 - 14:05

Yellow $= 5n - 1$

Red $= 10n - 9$

Blue $= n^2 + 5n$

Green $= n^2 + n$

 

 

Submitted by ISmile on Mon, 06/13/2016 - 14:35

It changes every time for me.
Eg: Level 1 - I put 123, it somehow lights up. I refresh the page, put in 123 and it doesn't light up.
I think, "That's weird." I refresh the page again, put in 123, and it lights up again...

Submitted by cfg21 on Fri, 06/24/2016 - 09:39

Yes, every time you refresh, a new set of rules will switch on the lights. It makes it more interesting!

Sometimes you're able to switch on all the lights at the same time...

Submitted by The Person on Wed, 06/22/2016 - 11:15

This is a strategy.
Think about each light in turn. Start at 1, then keep going up by one number at a time noting down the numbers that turn it on.
When you think you have a pattern, move on to the next light.
I hope this helps.

Submitted by Rory D , Solihu... on Fri, 06/24/2016 - 08:34

Red = 2n
Yellow =10n-6
Green = 3n-1
Blue = 9n-7

Submitted by cfg21 on Fri, 06/24/2016 - 09:42

Were you able to find four digit numbers that turned on each light?
Were you able to find any numbers that turned on more than one light?

Submitted by Katy Thomas on Fri, 06/24/2016 - 08:37

Yellow: even numbers
Red: 7n-3
Blue: 8n+1
Green: odd numbers

I'm from Solihull School

Submitted by Izzy Gardner on Fri, 06/24/2016 - 08:37

The yellow light is triggered by adding 5 each time starting from 2.
The red light is triggered by adding 7 each time starting on 3.
The blue one is triggered by adding 9 each time starting from 0.
The green one is triggered by adding 12 each time starting from 11.

A shorter way of writing these is:
yellow = 5n-3
red = 7n-4
blue = 9n
green = 12n-1

Submitted by James Gardner f... on Fri, 06/24/2016 - 08:42

Yellow = 5n-1
Red = 11n
Blue = 7n
Green = 3n
Solihull School

Submitted by Ataclib, Solihu... on Fri, 06/24/2016 - 08:42

Green: even numbers
Red: 6n-4
Blue: 11n-7
Yellow: 8n-1

Submitted by cfg21 on Fri, 06/24/2016 - 09:43

Do you think it would be possible to find a number that would light up all four lights at the same time?
If not all four, what about three lights? Or two lights?

Can you explain why some pairs of lights will never light up at the same time?

Submitted by magic maths on Tue, 06/28/2016 - 15:20

Has anyone worked out a formula for green light at level 3?
Our numbers were 3, 5, 7, 13, 17, 27, 33, 47, 55, 73, 83, 105, 117, 143....
Any ideas?

Submitted by cfg21 on Wed, 06/29/2016 - 13:16

Can you find a rule for your first, third, fifth, seventh... values: 3, 7, 17, 33, 55, 83, 117?

What happens if you substitute negative numbers into your rule?

Does it help to draw a graph of your quadratic to see what is going on?