Colour Wheels
Problem
Imagine a wheel with different coloured markings - red, green and blue painted on it at regular intervals.
As the wheel goes round, a trail is painted on the ground.
RGBRGBRGBRGBRGBRGB ... ...
Satisfy yourself that you can predict where the blues will appear.
How about the red and green marks?
Can you predict the colour of the 18th? 19th? 31st? 59th? 299th? 3311th? 96 312th?
How did you work it out?
Now consider the wheels that produce:
BBYGBBYGBBYG ... ...
BYBRBYBRBYBRBYBR ... ...
RRRBBYRRRBBYRRRBBYRRRBBYRRRBBY ... ...
RRBRRRRBRRRRBRRRRBRRRRBRR ... ...
What will the 24th colour be in each case? The 49th colour? The 100th?
How did you work it out?
You could continue this investigation by asking yourself some "what if ...?" questions:
A wheel has six markings. Where would red be painted on it so that the 100th mark made is red?
What other wheels (with more/fewer markings) would give you a red in the 100th position?
Getting Started
Can you describe the pattern?
How does the pattern repeat?
You could draw the sequence of colours and number each one. What do you notice?
Student Solutions
We had a good number of solutions sent in. Some sent in just the first part and that's fine - thank you. Others went through to the end.
Elijah who is home educated sent in the following.
RGBRGB wheelA multiple of $3$ is blue. (You can quickly see if a number is multiple of $3$ by adding its digits and seeing if that sum is a multiple of $3$.)
$1$ more than a multiple of $3$ is red.
$1$ less than (or $2$ more than) a multiple of $3$ is green.
So:
$18$th is blue
$19$th is red
$31$st is red
$59$th is green
$299$th is green
$3311$th is green
$96312$th is blue
BBYG wheel
A multiple of $4$ is green.
So $24$th is green.
$49$ is $1$ more than $48$, which is a multiple of $4$, so $49$th is blue.
$100$th is green.
BYBR wheel
A multiple of $4$ is red.
So $24$th is red.
$49$ is $1$ more than $48$, which is a multiple of $4$, so $49$th is blue.
$100$th is red.
RRRBBY wheel
A multiple of $6$ is yellow.
So $24$th is yellow.
$49$ is $1$ more than $48$, which is a multiple of $6$, so $49$th is red.
$100$ is one more than $99$.
$99$ is an odd multiple of $3$.
Odd multiples of $3$ are red.
Blue comes after red, so $100$th is blue.
RRBRR wheel
There are blue marks at these places:
$3, 8, 13, 18, 23, 28, 33, 38$ ....
If the number ends with $3$ or $8$ there will be a blue mark.
Everything else is red.
So $24$th, $49$th and $100$th are red.
A wheel with six markings
The $6$th colour marking will be on the multiples of $6$.
$102$ is a multiple of $6$.
$100$ is $2$ less than $102$, so red must be $2$ back from the $6$th marking.
So if you want the $100$th mark to be red, the red must be the $4$th colour marking on the wheel.
Other wheels to make a red $100$th mark
How many colour markings: Position of red colour marking:
$7$ ---------------------------------$2$nd
$8$ ---------------------------------$4$th
$9$ ---------------------------------$1$st
$10$ -------------------------------$10$th
Emma from Bradon Forest School in the U.K. Sent in a similar solution but these were her explanations for the RRBRR wheel.
RRBRR:
Because red appears on every $5$th colour, the pattern is $5, 10, 15, 20$.
The last red of every spin is in the $5$ times table.
As for the other colours you find out how many $5$s there are in the number and then add however many you need to get the number you want. If you need to add $1, 2$ or $4$ then the mark will be red but if you need to add $3$ then the mark will be blue.
E.g to find the $20$th colour you find out how many $5$s are in $20$ which is $4$. $4$ times $5$ = $20$. You don't need to add anything so the mark is red.
The $100$th mark: You divide $100$ by $6$ which = $16$r$4$. $16$ means the wheel would have to spin $16$ times to get to $100$. r$4$ means that red should be the $4$th spoke on it so it can be the $100$th mark.
Well done to the many of you for the ideas you shared and the solutions you sent in - keep them coming.
Teachers' Resources
Why do this problem?
Possible approach
Key questions
Possible extension