I have a somewhat quirky morning routine. I think it started as a way to occupy my mind during my morning commute. I don’t drive a car and so for the last 11 years of my teaching career, all spent in the same school, I used to use a combination of roughly 40 minutes walking and an eight-minute train journey to get to school each day. I like to be up for quite some time before I have to be active – to ease myself into the morning – and so I used to get up at 6 am, leave the house around 7:15, and arrive at school shortly after 8 am. During that time before I left the house, and then during the journey to school, I run through a series of games on my phone with an almost religious regularity.
It starts with the daily Wordle, then the daily Quordle
(both the Classic and the Sequence, as well as the Weekly when I solve my first
Classic of the week). Then I move onto the maths games, starting with the
Ooodle, the OoodleMax and the Time Square grid, followed by the Nerdle. Then it
is onto geography, with the Worldle (and its different rounds), the Statele, and
then the Globle. I finish off with the Daily Sudoku, and then have recently
added the Countle at the end. Since leaving teaching I have continued to play
these games as part of my morning routine before starting my work from home.
The reason I mention all of this is that my attempt at the OoodleMax
today prompted me to consider a nice mathematical relationship, that would make
an excellent prompt for learners that could be used at multiple levels.
This was today’s Ooodle Max:
The goal is to use the numbers from the keypad on the right-hand side, at most once each, to fill in the blanks and make the target. Like
other Wordle type games, numbers go orange if they appear in the calculation,
but are in the wrong place, green if they are in the correct place in the
calculation, and grey if they are not used at all. This was my first guess:
Having used 12 and 15, and without really thinking about it, it occurred
to me that to be close I should use 13 and 14 in the multiplication. I then
knew the “2” would have to be the divisor of the division, so I then set about
figuring about what the dividend would be. This was my second guess:
I was lucky in that I got the 13 and 14 the right way, but
what struck me is the relationship between the value of 13 × 14 and 12 × 15. It
reminded me of the result that I knew about, which is that the square of an
integer is always one more than the product of its adjacent integers. This is,
of course, easy to prove mathematically for learners that can expand binomials.
Given three integers 
then we have that 
. It got me thinking
that there must be a wider pattern to this, which of course there is; given
four numbers 
, the product 
and 
(so a difference of 2), and then given 
we have 
and 
(a difference of 3) and so on. There is an
obvious symmetry here, as we get further away from a central value the difference
between the products increases by one and starting the list each time with 
makes this relatively clear each time.
The algebraic exploration of this is
clearly a nice activity for those that can access the algebra, particularly
proving the overall general case. However, it also struck me that these sorts
of related calculations, and the general structure underpinning the relationship,
can be explored without the need for algebra, using concrete manipulatives:
These are representations of 2 × 3 and 1
× 4, and then 4 × 5 and 3 × 6 – following the same pattern as 13 × 14 and 12 ×
15. The question then becomes how is the red array related to the yellow? It is
a relatively simple matter to see that we can move counters from the bottom row
of the yellow arrays and attach them to the side to create the red arrays, but
we will always have two counters left over.
It should also be within the capabilities
of many learners to reason why this must always be the case; the number of rows
in the red array is two less than the number of columns in the (original)
yellow array, and so when we move counters from the bottom row of the yellow
array to the side to extend the size of each row, we will always leave two
behind. This sort of argument can then be generalised further to any of the
related calculations.
So, what would the task look like? I haven’t
decided fully yet, but as an inquiry prompt it might be something like this:
Calculate the following:
(a)
2 × 3 and 1 × 4
(b)
3 × 4 and 2 × 5
(c)
4 × 5 and 3 × 6
(d)
13 × 14 and 12 × 15
What do you notice about each pair of results?
Could you write down other pairs that
would produce the same result?
Can you explain it?
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