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# Christmas Chocolates

Another way to work it out would be to add the rows up.

You can begin to see a pattern; the hexagonal box can always be divided into six triangles, or three parallelograms, or can be stripped down layer by layer like Matthew did. This is true whatever size the box is (as long as it has six sides). This important characteristic can then be used as a starting point for developing a general formula for the number of chocolates in the box. Philip from Gosforth High School used the pattern:

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This problem demonstrates that multiple, different methods can be used to approach a particular question. In addition, you are encouraged to devise a general formula for the number of chocolates in any hexagonal box. In doing this, it then becomes very straightforward to work out the number of chocolates there are, however large the box is.

Working out general formulae is not only fun and satisfying, but is also very useful; it provides a quick way to extend the original set of data, without having to work out every consecutive point. Scientists, Mathematicians, Engineers, Accountants...they all find general formulae very helpful, and make use of them regularly.

In the Christmas Chocolates problem, the formula can be extended to any size of box, because the differences are uniformly spaced. However, sometimes we do not have "perfect" information. For example, an experiment may yield slightly different results each time (this is why we do repeat experiments), but we cannot do an infinite number of trials. Thus we settle on an appropriate number, and then use the data collected to see if there is a pattern. From this, perhaps with more experiments, we may be able to devise a general formula. This can then be used to make predictions, and then experiments can be used to test these. The formula may then be refined, and then new predictions made, and so on. Remember though that the formula may not be true for all conditions, unlike the chocolate problem. For example, the formula may only apply within a given temperature range. Nevertheless, having a formula is very useful; it not only makes calculations more efficient, but can also yield great insight into the properties of the problem.

Lots of great solutions were submitted, so thank you to everyone for this! Many students noticed the shapes that were made once Penny, Tom, and Matthew had eaten some chocolates. They then realised that several of these shapes could fit into the whole hexagonal box. This is useful as it provides a convenient way of grouping the chocolates together, and therefore quickens calculations. The patterns can then be used to extend to other box sizes, as several students did.

Courtney, from Penrhos College, submitted this solution. It is lovely: clear, and illustrated. Lauren, from Tudhoe Grange, also submitted a good solution. This solution is colourful, and demonstrates the different shapes created by Penny, Tom, and Matthew very clearly. Adam, from Wilson's School explained:

Penny worked out that there were $61$ chocolates by using
triangles of $10$ chocolates. Penny would have found that she could
find six of these triangles. $6\times10=60$, but there is still the
chocolate in the middle of the box, so we add this on, which makes
$61$.

Tom would have used parallelograms of $20$ chocolates; these
parallelograms are "four by five". He could find three of these
parallelograms. $3\times20=60$, but again there is still the
chocolate in the centre so there would be $61$ chocolates in
total.

Matthew ate all the chocolates on the outside, which makes
$24$ chocolates. As there are 24 chocolates on the outside it must
be $(5-1)\times6$. This is because there are six sides, and if you
multiplied six by five you would count six chocolates twice. Each
layer, moving inwards, decreases by six. So, to work out how many
chocolates there are in the box the sum is:

$[(5-1)\times6]+ [(4-1)\times6]+ [(3-1)\times6]+
[(2-1)\times6]+1$, which equals 61.

Another way to work it out would be to add the rows up.

Luke, from Wilmslow High School, used the alternative method suggested by Adam. In his solution, he added the rows of the hexagon up to find out the total number of chocolates in the box.

Monty, from The Perse School, and Claudia & Martha from Impington Village College noticed that Penny and Tom's methods were extremely similar; Tom ate twice as much as Penny, so the only difference between the caluclation is a factor of two!

Philip, from Wilson's School extended the methods used by Penny, Tom, and Matthew to work out the number of chocolates in a box with ten along each edge:

If the sides of the box measure $10$ chocolates:

Penny would find out that the box would have $271$ chocolates
by realising that there were six triangles made up of $45$
chocolates each, plus one extra chocolate in the middle.

Tom would work out that there were $271$ chocolates in the box
because he would realise that the box can be divided into three
parallelograms, made up of $90$ chocolates each, plus one extra
chocolate.

Matthew would work out that there were $271$ chocolates in the
box because he would realise that the outer layer of chocolates
would be made up of $54$ chocolates and each layer afterwards would
have a decrease of $6$ chocolates, plus one extra chocolate in the
middle: $54+48+42+36+30+24+18+12+6+1 = 271$

You can begin to see a pattern; the hexagonal box can always be divided into six triangles, or three parallelograms, or can be stripped down layer by layer like Matthew did. This is true whatever size the box is (as long as it has six sides). This important characteristic can then be used as a starting point for developing a general formula for the number of chocolates in the box. Philip from Gosforth High School used the pattern:

If each child was given a chocolate box with $10$ along each
side, the number of chocolates in the box would be $271$. This
would be because the number along the bottom of the triangle (like
Penny created) is always one
less than the number along each side, therefore the base of
the triangle would be nine chocolates wide. This would also make
the number going up the side nine, which in turn would make each
triangle have $45$ chocolates in it (the triangle number of nine).
As there is always one chocolate which is not inlcuded in any of
the triangles, you add this to the sum of $45\times6$ ($6$= the
number of these triangles in the hexagon) to get $271$.

Following on from this, several people submitted correct
formulae for the number of chocolates in a box with n along each
side. These include: Adam, Philip, and Elliot from Wilson's School,
Thomas from Perse Upper School, Rajeev from Fair Field Junior
School, and Hannah & Phil (they did not submit a school).
Hannah and Phil explained how they worked out a formula:

We felt that Penny's method was most effective (for us), so we
tried solving for the quantity of chocolates of a hexagonal box,
with a side length of 10 chocolates. We tried coming of up with a
formula to calculate the number of chocolates, using Penny's
method.

So we went back and analysed the equation $(10\times6)+1=61$,
and soon realized those numbers represented this: [(Number of
chocolates per triangle) x 6] + 1 = total number of chocolates.
Then we tried finding a formula for how to achieve the number of
chocolates per triangle. Soon we realized that it consisted of
rows, each one less than the one underneath:

1 O

2 OO

3 OOO

4 OOOO

5 OOOOO

This reminded us of previous problems we have dealt with based
on this principle, such as finding the sum of the numbers from 1
through 10. The formula was: $( ½n)\times(n-1)$.

Then we tested this formula to see if it applied for finding
the number of chocolates per triangle:
$( ½n)\times(n-1)$

Substitute n=$5$ into the formula:
$( ½(5)\times(5-1))= 2.5\times4= 10$.

The formula did work.

So then we were able to put the whole formula together, for
the whole box of chocolates, based on Penny's method.

We had: $[( ½n)\times(n-1)\times6]+ 1$. This then
simplified to: $[3n(n-1)] +1$.

Now we could solve the problem with a hexagon of side length
10 chocolates:

$[3n (n-1)] +1$

Substitute n=$10$ into the formula:

$[3(10)\times(10-1)] +1= (30\times9) +1= 270 + 1= 271$.

A box of chocolates with side length 10 would have 271
chocolates inside. When analysing this formula, we realized, that,
unfortunately, there is no size the chocolate box can be that would
allow the three children to share the chocolates among themselves
equally. The number of chocolates would have to be divisible by
$3$, and the formula shows that the number of chocolates will
always be $1$ more than a multiple of $3$.

Kevin, Leif, Oliver, Nisha, Olivia, Thomas, and Henry from Highgate School explained the latter point made by Hannah and Phil (about sharing the chocolates equally). They used Matthew's method as an example:

To find the number of chocolates in one layer (like Matthew
did) you multiply the number of chocolates (minus one) on each side
by six, which means that this number is a multiple of six. It is
therefore also a multiple of three. Every layer surrounding the
middle chocolate is also a multiple of six (and therefore of
three), so when you share out the chocolates between the three of
them the last one (the one in the middle) will be left over.

So there is no size of box which means that the three of them
can share the chocolates out equally.

They then applied the same principle to work out the size of chocolate box the boys will be able to share out equally (i.e. between two people):

The number of chocolates in each layer (here we are referring
to the layers created by Matthew) is a multiple of six, and so it
is also an even number. So, when you add all the layers together
and then add the middle chocolate, the total will be an odd number. So there is no size
of box the two boys can share out equally.

In fact, as Philip from Wilson's School pointed out:

Whatever the number of chocolates in the box, the children
will never be able to share them out equally.

What's the moral of this, then? Well, if you want to share a box of chocolates equally between two or three people, it should not be a hexagonal box like the one in the question. You could, of course, not share it equally... and have more for yourself... Even better, give more to somebody else!

Thank you to everyone who submitted solutions to this tasty chocolatey problem. If you enjoyed this (and we know that you did!), have a go at Summing Squares and Picture Story. These problems are nice extensions of this chocolate problem. If you would like to try a similar, related problem for more practice, try Picturing Triangle Numbers, Mystic Rose, or Handshakes.

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?

Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?

The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?