Tuesday 28 November 2017

On 'Incredible' Feats of Mental Arithmetic (in Countdown)









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I was piqued by this tweet and the image below from @QEHSmaths, showing an inattentive (mis)use of the equals sign by Carol Vorderman on an old episode of Countdown [1].  After getting over some of my initial bemusement, and clearly not being overly bothered by the rough setting-out, I was most struck by the problem itself, and specifically by the two solutions on the board, where at least one contestant had made 840, and the other contestant (or Carol) had made the 833 required.

I was struck by this because the two solutions respectively typify approaches adopted by 'winners' in the numbers game and 'losers'.  In other words, one approach marks out a strategy that often leads to a solution in the game, whilst the other illustrates a strategy that, in the time given, is limiting.  I reflected on how contestants are expected to relay back their solutions in the game (step-by-step, providing intermediary answers), and on how we as teachers, inadvertently or not, may relay similar expectations of our students in our classrooms in all manner of ways, not least in how we place calculation first, systemically, above the thinking that underpins it — above, that is, computational thinking, as espoused by, amongst others, Conrad Wolfram (see here as one example, and listen to this thought-provoking interview).  I thought about the times I've seen the Countdown numbers game used in secondary classrooms, more often than not simply as is, usually with younger students, and the opportunities that we sometimes miss to develop our students' mathematical maturity by just not going slower, by not delving deeper into the core of the problems themselves.

To this end, I have collated here a handful of clips of 'incredible' solutions from the pantheon of Countdown number game moments, for teachers to use with students — not in an effort to engage them with mental calculation, but rather to engage them in an evaluative exercise that will elicit an appreciation of elegance, efficiency and, indeed, the power of algorithmic approaches to computation.  The clips are intended to pique students' curiosities enough for our starting questions to have a greater chance to encourage deeper learning.  After watching the clips with students [2], teachers should elicit students' initial reactions and responses, with a simple 'what do you think?' if necessary.  Some starting questions are also offered as suggestions.  (Clicking on the link at the top of each clip will open the video outside of this blog.)


The Game

Firstly, in case you are not aware of the Countdown 'numbers' problem: A contestant chooses a problem set (p) of six numbers from a set (a) of four 'large' numbers and a set (b) of 20 'small' numbers:
\[\begin{array}{l}\begin{align}a &= \left\{ {25,50,75,100} \right\}\\b &= \left\{ {1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10} \right\}\end{align}\end{array}\]
A 3-digit target number {t | ℕ, 99 < t < 1000} is then randomly generated, which the contestants must try and make using only the four basic arithmetic operations and using the numbers in only once — but not necessarily all of them.

Some facts / questions that may help to alter students' perceptions of the game, positioning them away from the mere arithmetical challenge and more towards a mathematical way of thinking:
  • There is only one set p from which it is impossible to make any t.  
  • What is it and why are there no others?
  • The most difficult t to make is 947.  
  • What do we mean by saying it is the most difficult, and how do we know?
  • There are 1226 sets of numbers from which all target numbers can be made (for the full list of sets see here).  In other words, if you take one of these sets, say {1,2,7,9,10,50}, you can make all the numbers from 100-999 (see here for a full list of results). [3][4]
  • There are 13,243 possible combinations of six tiles.  Show that this is the case.


Some possible starting questions for students to consider (or to consider with students) after watching the clip:
  • Could James have solved the problem in a different way?
  • Did Carol need to find the answer to every step?
  • The contestant gives us a clue that he may not have found the answer to every step in the solution (1:55 in), so how then did he know the result?
  • What was Carol trying to explain (3:04 in)?

The 'incredible' — in the words of Carol ('remarkable' in the words of presenter Richard Whiteley) — solution, in the step-by-step form that the program uses, is as shown:

\[\begin{array}{c}\begin{align}\left( {100 + 6} \right) \times 3 &= 318\\&\Rightarrow318 \times 75 = 23,850\\&\Rightarrow23,850 - 50 = 23,800\\&\Rightarrow\frac{{23,800}}{{25}} = 952\end{align}\end{array}\]

The solution broken down without showing intermediate steps, as it is probable that James Martin used, is thus:

\[\begin{array}{c}\begin{align}\frac{{75\left[ {3\left( {100 + 6} \right)} \right] - 50}}{{25}} &= \frac{{\mathop {\rlap{-} 7\rlap{-} 5}\limits^3 \left[ {3\left( {100 + 6} \right)} \right]}}{{\mathop {\rlap{-} 2\rlap{-} 5}\limits_1 }} - \frac{{\mathop {\rlap{-} 5\rlap{-} 0}\limits^2 }}{{\mathop {\rlap{-} 2\rlap{-} 5}\limits_1 }}\\ &= 3 \times 3\left( {100 + 6} \right) - 2\\ &= 9 \times 106 - 2\\ &= 954 - 2\\ &= 952\end{align}\end{array}\]


Sam's 821 (Letters and Numbers, Australia, 2012).
Some possible starting questions for students to consider (or to consider with students), after watching the clip:
  • Which contestant's solution most impressed you?  Why?
  • What does Sam's solution suggest about his way of thinking?
  • Why do you think of the presenter commended the 'sheer audacity' of the solution?
  • What do you think about Lily’s working out?

The 'amazing' — in the words of Richard Morecroft the presenter — 'remarkable' solution, in the step-by-step form that the program uses, is as shown:

\[\begin{array}{c}\begin{align}\left( {50 \times 4} \right) + 6 &= 206\\&\Rightarrow206 \times 100 = 20,600\\&\Rightarrow20,600 - 75 = 20,525\\&\Rightarrow\frac{{20,525}}{{25}} = 821\end{align}\end{array}\]
The solution broken down without showing intermediate steps, is thus:

\[\begin{array}{c}\begin{align}\frac{{100\left[ {4\left( {50} \right) + 6} \right] - 75}}{{25}} &= \frac{{\mathop {\rlap{-} 1\rlap{-} 00}\limits^4 \left[ {4\left( {50} \right) + 6} \right]}}{{\mathop {\rlap{-} 25}\limits_1 }} - \frac{{\mathop {\rlap{-} 75}\limits^3 }}{{\mathop {\rlap{-} 25}\limits_1 }}\\ &= 4 \times \left( {200 + 6} \right) - 3\\ &= 4 \times 206 - 3\\ &= 824 - 3\\ &= 821\end{align}\end{array}\]
The solution offered by the other contestant was this:

\[\begin{array}{c}\begin{align}6\left( {100 + 25} \right) + 75 - 4 &= \left( {6 \times 125} \right) + 71\\ &= 750 + 71\\ &= 821\end{align}\end{array}\]


The John O'Neill 813 game (Countdown, UK, 2002). [5]
= {1,10,25,50,75,100}, = 813.



Some possible starting questions for students to consider (or to consider with students), after watching the clip:
  • What do you think about this solution in comparison with the previous two clips?
  • Why does the calculation look impossible from the start?
  • Are you noticing similarities in the respective contestant's strategies?

The 'really clever' — in the words of Carol Vorderman ('fantastic' in the words of presenter Richard Whiteley) — solution, in the step-by-step form that the program uses, is as shown:

\[\begin{array}{l}\begin{align}75 - 10 &= 65\\ &\Rightarrow 65 \times 25 = 1625\\ &\Rightarrow 1625 + 1 = 1626\\ &\Rightarrow 1626 \times 50 = 81,300\\ &\Rightarrow \frac{{81,300}}{{100}} = 813\end{align}\end{array}\]
Students could be asked to show the solution broken down without showing intermediate steps, and then compare this with the step-by-step solution in terms of computational efficiency.  They may notice some difference between this and the previous examples.


Noel McIlvenny 546 and 965 games (Countdown, UK, 2017).
p = {4,5,25,50,75,100}, t = 965



Some possible starting questions for students to consider (or to consider with students), after watching the clip:
  • In the first numbers game, which contestant’s solution do you find more elegant?  Why?
  • In the first numbers game, what does the difference in the two contestants' solutions suggest about their relative ways of thinking?
  • Could the 2nd numbers game be solved in a different way?
  • What strategy is Noel using, that contestants in previous clips also seem to be using?
  • Is there a 'fast' way of mentally dividing by 25? 

Noel's solution to the first numbers game, in the step-by-step form that the program uses, is as shown:

\[\begin{array}{l}\begin{align}75 - 7 &= 68\\ &\Rightarrow 68 \times 100 = 6800\\ &\Rightarrow 6800 \times 2 = 13,600\\ &\Rightarrow 13,600 + 50 = 13,650\\ &\Rightarrow \frac{{13,650}}{{25}} = 546\end{align}\end{array}\]
Students could be asked to show this solution, and the solution in the second numbers game, broken down without showing intermediate steps, and then compare them with the respective step-by-step solution in terms of computational efficiency, nudging students in the direction of computational thinking.

It is worth noting that all of the examples collated above, all of those number games in other words that seem to astound people, are those when the contestant chooses the maximum allowable number of ‘large’ numbers (i.e. all four: 25, 50, 75 and 100).  Choosing the four large numbers reduces the set of numbers that are likely to be drawn, so if you are a contestant who chooses four large numbers, it is given that the numbers 25, 50, 75 and 100 will be in p, meaning that the remaining two numbers will come from a set of ten.  But, students could be asked, does this help you or hinder you, does it help you by hindering your opponent?  Some further possible questions for students to consider (or to consider with students):
  • Does choosing the numbers 25, 50, 75 and 100 from the six really make the problem more challenging?  
  • What is it about the ‘four large and two small’ option that makes, or seems to make, incredible feats of mental arithmetic?
  • Why do you think most contestants choose one large and five small numbers?
  • Is it better to take 0,1,2,3 or 4 large numbers?
  • You get ten points if you solve the number puzzle.  How many large numbers would you choose if you were eleven points ahead with one round to go?  Why?
  • Would it be possible to have such ‘astounding’ solutions if contestants chose all ‘small’ numbers?  (See the clip that follows.)  


Dylan Taylor small solve (Countdown, UK, 2013).
= {1,2,3,5,7,9} t = 924




Some possible twists to the Countdown number game:
  • Write the solution out formally in one single line equation, then simplify as much as possible.
  • Find the result using all numbers, or — try and — prove it if you cannot.
  • Find the result using numbers in lexicographic order.
  • Find as many ways of getting the result as possible.
  • Find the result using the least amount of numbers.
  • You are allowed to concatenate numbers, i.e. combine two or more numbers as if they were digits, i.e. taking 8 and 3 as an example, 8||3 or  3||8 to give 83 or 38.
  • You are allowed to use any of the numbers as an index.

There are various free online applets that you can use for the Countdown numbers game (for example this via NRICH, who have also developed one using fractions).  There are also a range of Number Game Solvers to be found online, giving you all possible solutions to the problem (for example this from incoherency.com, this from datagenetics.com, and this from crosswordtools.com).



Notes & (Select) Links:

[1] Countdown is a British game show that tests its contestants vocabulary and arithmetic skills.  It has been running for 35 years, and is the British version of Des Chiffres et des lettres

[2] Whilst students may of course compete and solve the problems themselves, at the same time as the contestants, it is important to note that this is fundamentally not the aim of the exercise.  Competing in 'real time' should have the effect of hooking students in to what is planned to follow.

[3] For an in-depth numerical analysis of the Countdown numbers game, see this brilliant blog post by datagenetics.com — something that may indeed be worth sharing with students.

[4] Similarly, this paper may be illuminating to share with students, in the sense that it illustrates the  mathematics underpinning an algorithmic approach to the problem.

[5] See this for more on John O'Neill. 


1 comment:

  1. Didn't check over all of it, but you've got a mistake:
    "The 'amazing' — in the words of Richard Morecroft the presenter — 'remarkable' solution, in the step-by-step form that the program uses, is as shown:

    (50×4)+6=246⇒246×100=24,600⇒24,600−75=24,525⇒24,525/25=821"

    (50x4)+6 should of course be 206, not 246. That end bit needs to be 20525/25 = 821... the written 24525/25 is 981.

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