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Exploring Wild & Wonderful Number Patterns

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4 Dom

Stage: 1, 2, 3 and 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Well done to all of you who found a solution to this challenge. Chloe and Mia from Scotts Primary School both managed to solve this challenge whilst in competition with Mrs Briggs, the teaching assistant. They both finished really quickly apparently and were really pleased with themselves (quite rightly!). Their teacher, Miss Lilley, and the rest of the class were very proud of them!

Zareah from St Joseph's Convent gave us some good general advice:

You just need to even it out. Don't put all the larger numbers in one corner and all the smaller numbers in another corner. Put a small number with a large number and soon you'll get it out!

Milo and Darren from SVS used a similar method to Zareah. Angus from West Linton Primary used this strategy for solving the problem:

First I added up the dots ($5+1+6+4+2+6+4+3=31$) then divided it by $4$ to find the average dots on a domino.
There are one and a half dominoes in a side.
$31\div4 =8$ (roughly)
$8\times 1\frac{1}{2}= 12$
Then I fitted $12$ domino spots into each side.

What a good idea, Angus. This estimation really helped you! Here is Angus' picture of the solution:




Lisa, who didn't give her school, described a very convincing way of working out the solution:



This is the lowest side total possible using the domino with the highest total: $6+4+1=11$.
The sides can therefore not add up to less then $11$.
However, for the $5+1$ domino to be part of a side totalling $11$, there would need to be another $5$. Because there is not another $5$ the lowest possible total of the sides is $12$.



This is the highest total possible using the domino with the lowest total: $5+1+6=12$.
Combining this with the information we already have, we know the sides must add up to $12$.
It is now simply the case of working out the number needed to make $12$.







$6+4+2=12$ There is only one $2$, so this domino must go here.


$6+2+4=12$ The remaining domino must therefore be placed this way round.


$3+4+5=12$ If the $5+1$ domino was the other way round it would need to be rotated to make the final side add up.

Very well explained, Lisa. We can see that this is the same solution as Angus', just rotated. A pupil from Wootton Upper School had a slightly different method:

Part 1: How many dots on each side?
The arrangement shows that each side of the square is composed of the sum of both sides of a single domino plus one of the sides of another domino.
The sum of the dots on each single domino is as follows: ($5$,$1$) = $6$, ($6$,$4$) = $10$, ($4$,$3$) = $7$ and ($6$,$2$) = $8$.
Therefore, the smallest sum possible is $8$ i.e. $6$ (from $5$,$1$) + $2$ (from $6$,$2$) whereas the largest sum possible is $16$ i.e. $10$ (from $6$,$4$) + $6$ (from $5$,$1$).

She then conjectured that, since the four sides of the square must have equal numbers of dots, a compromise of the extreme values is needed so each side could total $12$ dots.

Part 2: Arrangement
The ($5$,$1$) domino needs a $6$ from one side of another domino since $6 + 6 = 12$
Similarly, ($6$,$4$) needs a $2$ as $10 + 2 = 12$
($4$,$3$) needs a $5$ as $7 + 5 = 12$
($6$,$2$) needs as $4$ as $8 + 4 = 12$

For $6 + 6$, the $6$ can be obtained from ($6$,$4$) or ($6$,$2$)
For $10 + 2$, the $2$ can only be obtained from ($6$,$2$)
For $7 + 5$, the $5$ can only be obtained from ($5$,$1$)
For $8 + 4$, the $4$ can be obtained from ($6$,$4$) or ($4$,$3$)

Looking back at the square, it should be acknowledged that the dominoes are connected from head to tail like a chain. This is what we need to do with our numbers - arrange them so they form a nice chain.

For the above reason, we use ($6$,$4$) to obtain the $6$, and ($4$,$3$) to obtain the $4$ so that each domino is linked up consistently and they all get their required number of dots to total $12$.

The result obtained should be: ($6$,$4$) - ($2$,$6$) - ($4$,$3$) - ($5$,$1$) - ($6$,$4$) etc.