Jørgensen.org.uk - Thoughts/2018-09-26T00:50:55+01:00Homework for a 9-year-old2018-09-26T00:40:25+01:002018-09-26T00:50:55+01:00Karl E. Jørgensentag:None,2018-09-26:/homework-for-a-9-year-old.html<p>A <a href="https://www.facebook.com/groups/stanfordlehope/permalink/1869988459703153/">post about some maths
homework</a>
caused some consternation and confusion on one of my local facebook groups:</p>
<p><img alt="homework" src="/Thoughts/homework-small.jpg"></p>
<p>Lots of people seemed to think this was far too advanced for a
9-year-old, or that solving such puzzles would have no applications in
real life.</p>
<p>I beg to disagree.</p>
<p>Yes, I …</p><p>A <a href="https://www.facebook.com/groups/stanfordlehope/permalink/1869988459703153/">post about some maths
homework</a>
caused some consternation and confusion on one of my local facebook groups:</p>
<p><img alt="homework" src="/Thoughts/homework-small.jpg"></p>
<p>Lots of people seemed to think this was far too advanced for a
9-year-old, or that solving such puzzles would have no applications in
real life.</p>
<p>I beg to disagree.</p>
<p>Yes, I admit that solving the puzzles takes a sizeable number of
steps, and might test the concentration of your little ones, but the
steps themselves are easy to understand and easy to execute. They <em>do</em>
require a decent understanding of numbers though.</p>
<p>The underlying principle of solving <em>any</em> maths equation is to use the
known facts to derive other things which we can treat as new facts.
And then using <em>those</em> facts to derive even more stuff.</p>
<p>In other words: Common sense. You can formalise this into
<a href="https://en.wikipedia.org/wiki/Propositional_calculus">propositional
calculus</a>,
<a href="https://en.wikipedia.org/wiki/First-order_logic">first-order logic</a>
or whatever you like, but at the end of the day <em>it is common
sense</em>. Or rather: common sense on stereoids.</p>
<p>In this sense, the puzzles presented to the 9-year-olds are less about
formal maths than general problem solving: Using what you already know
to figure out stuff. Which then allows you to figure out <em>more</em>
stuff. Rinse and repeat ad nauseum. Surely this is something an
education system is meant to do!</p>
<p>It was presented on facebook without much context, but there are some
reasonable assumptions to make here:</p>
<ul>
<li>
<p>each letter represents a single digit</p>
</li>
<li>
<p>two-digit numbers are represented as two letters. This is
contrary to the normal maths notation where there would be an
implicit multiplication</p>
</li>
<li>
<p>In each table, each letter stands for a <em>different</em> digit. No two
letters stand for the same digit within the same table.</p>
</li>
<li>
<p>With only 9 distinct letters in each table (except for Table Y),
some digit must be left out - we do not know which one, but zero
sounds like a reasonable guess. Nobody likes zeros anyway. They
don't matter.</p>
</li>
<li>
<p>The goal of the puzzle is to figure out which letters represents
which digits</p>
</li>
</ul>
<p>A quick refresher on notation:</p>
<ul>
<li>
<p>The <span class="math">\(\Rightarrow\)</span> symbol should be read as "implies that". So <span class="math">\(A
\Rightarrow B\)</span> can be read as "A implies B": If A is true, then B
is also true. And if A is false, then B is also false.</p>
</li>
<li>
<p><span class="math">\(\times\)</span> is the multiplication symbol. But you did know that, right?</p>
</li>
</ul>
<h1>Table W</h1>
<p>This is the easiest one to solve - it's basically a taster:</p>
<p><img alt="Table W" src="/Thoughts/table-w.jpg"></p>
<p>With the assumptions above, we can focus in on <span class="math">\(H \times H = H\)</span> : this
means <span class="math">\(H\)</span> can only be zero or one, no other numbers are possible. And
since e.g. <span class="math">\(H \times A = A\)</span>, this means that <span class="math">\(H\)</span> cannot be zero. Hence <span class="math">\(H = 1\)</span>.</p>
<p>And that's it! The remaining letters can represent whatever digits
you like (including zero), and the equations will all be true.</p>
<h1>Table X</h1>
<p>Here we step up the difficulty level slightly:</p>
<p><img alt="Table X" src="/Thoughts/table-x.jpg"></p>
<p>Yes: the authors skipped the letter <em>E</em> for some reason....</p>
<p>From <span class="math">\(D \times A = D\)</span> we can conclude that <span class="math">\(A = 1\)</span>.</p>
<p>And considering <span class="math">\(D \times D = JA \Rightarrow D \times D = J1\)</span> : So <span class="math">\(D\)</span>
times itself must result in a two-digit number whose last digit is
<span class="math">\(1\)</span> : The only possiblity here is <span class="math">\(D=9, J=8\)</span></p>
<p>Plugging in those values into the remaining equations revals the
value of some more letters:</p>
<p><span class="math">\(D \times I = AJ \Rightarrow 9 \times I = 18 \Rightarrow I = 2\)</span></p>
<p><span class="math">\(D \times J = HI \Rightarrow 9 \times 8 = H2 \Rightarrow H = 7\)</span></p>
<p>We can then use the newly found value of <span class="math">\(I\)</span> and <span class="math">\(H\)</span> in one of the
other equations:</p>
<p><span class="math">\(D \times C = IH \Rightarrow 9 \times C = 27 \Rightarrow C = 3\)</span></p>
<p><span class="math">\(D \times H = FC \Rightarrow 9 \times 7 = F3 \Rightarrow F = 6\)</span></p>
<p><span class="math">\(D \times F = BG \Rightarrow 9 \times 6 = BG \Rightarrow B=5, G=4\)</span></p>
<p>And we then know the values of all the letters - even though we
haven't "used" all the equations yet.</p>
<h1>Table Y</h1>
<p>Here we step up the difficulty level even futher. We have 10 letters,
but still only 9 equations:</p>
<p><img alt="Table Y" src="/Thoughts/table-y.jpg"></p>
<p>From <span class="math">\(C \times B = C\)</span> we can conclude that <span class="math">\(B = 1\)</span>. Yes, the devious
teachers <em>hid it</em> in the middle, rather than presenting it as the
first equation! They expect you to understand the principles by now!</p>
<p>But then it gets slightly trickier...</p>
<p>Consider <span class="math">\(C \times C = AC\)</span> : The square of <span class="math">\(C\)</span> is a two-digit number
with <span class="math">\(C\)</span> as its last digit. Hence <span class="math">\(C = 5\)</span> or <span class="math">\(C = 6\)</span>, but we do not
know which one yet.</p>
<p>If we were to guess at <span class="math">\(C = 5\)</span>, then the last digit of any of the
products must be either <span class="math">\(0\)</span> or <span class="math">\(5\)</span> - which is not the case (we would
only see two possible values as the right-most digit on the right-hand
side of the equations, but there are obviously more than that). Hence,
<span class="math">\(C\)</span> cannot be <span class="math">\(5\)</span> after all. So it must be the case that <span class="math">\(C = 6\)</span>.</p>
<p>If <span class="math">\(C = 6\)</span> and <span class="math">\(C \times C = AC\)</span> then <span class="math">\(A = 3\)</span>.</p>
<p>As before, we can take the values we know so far and plug into one of
the remaining equations to develop more knowledge:</p>
<p><span class="math">\(C \times A = BE \Rightarrow 6 \times 3 = BE \Rightarrow 6 \times 3 = 18 \Rightarrow B=1, E=8\)</span></p>
<p><span class="math">\(C \times E = DE \Rightarrow 6 \times 8 = D8 \Rightarrow D = 4\)</span></p>
<p><span class="math">\(C \times D = GD \Rightarrow 6 \times 4 = G4 \Rightarrow G = 2\)</span></p>
<p><span class="math">\(C \times I = DG \Rightarrow 6 \times I = 42 \Rightarrow I = 7\)</span></p>
<p>which leaves F, H and J as unknowns and no letters (yet) representing
0, 5 or 9. And these two equations are yet to be solved:</p>
<p><span class="math">\(C \times J = AF \Rightarrow 6 \times J = 3F\)</span> : So <span class="math">\(J\)</span> has to be either <span class="math">\(5\)</span> or <span class="math">\(6\)</span> making <span class="math">\(F\)</span> either <span class="math">\(0\)</span> or <span class="math">\(6\)</span>. But we already have a letter for <span class="math">\(6\)</span> (<span class="math">\(C\)</span>), which leaves only the possibility of <span class="math">\(F = 0\)</span> which implies that <span class="math">\(J = 5\)</span>.</p>
<p>Which leaves the last equation:</p>
<p><span class="math">\(C \times H = JD \Rightarrow 6 \times H = 54 \Rightarrow H = 9\)</span></p>
<h1>Table Z</h1>
<p>In this table we're back to 9 letters (for some reason the authors
decided to miss out on the letter E again!??) and 9 equations:</p>
<p><img alt="Table Z" src="/Thoughts/table-z.jpg"></p>
<p>As before, from <span class="math">\(H \times G = H\)</span> we can conclude that <span class="math">\(G = 1\)</span>.</p>
<p>Plugging <span class="math">\(G = 1\)</span> into the first equation gives:</p>
<div class="math">$$H \times J = GJ \Rightarrow H \times J = 10 + J$$</div>
<p>... after subtracting <span class="math">\(J\)</span> from both sides of the equation:</p>
<div class="math">$$(H-1) \times J = 10$$</div>
<p>And since the only factors of 10 is 2 and 5, then one of the following
must be true:</p>
<ul>
<li><span class="math">\((H-1) = 2\)</span> and <span class="math">\(J = 5\)</span>, i.e. <span class="math">\(H = 3\)</span> and <span class="math">\(J = 5\)</span></li>
<li>or <span class="math">\((H-1) = 5\)</span> and <span class="math">\(J = 2\)</span>, i.e. <span class="math">\(H = 6\)</span> and <span class="math">\(J = 2\)</span></li>
</ul>
<p>... but we do not yet know which one.</p>
<p>We also have the equation <span class="math">\(H \times H = F\)</span> - so <span class="math">\(H\)</span> must be either 0,
2 or 3 (it cannot be 1, since <span class="math">\(G = 1\)</span>). Any higher numbers would
result in <span class="math">\(H \times H\)</span> ending up as a two-digit number. And then we
can exclude <span class="math">\(H = 0\)</span>, as then we should have expected an <span class="math">\(H\)</span> on the
right-hand side of the equation.</p>
<p>So it <em>must</em> be the case that <span class="math">\(H = 3\)</span>, as this is the only value which
can satisfy <em>both</em> the previous 2 paragraphs!</p>
<p>And since <span class="math">\((H-1) \times J = 10\)</span>, then <span class="math">\(J = 5\)</span></p>
<p><span class="math">\(H \times H = F \Rightarrow 3 \times 3 = F \Rightarrow F = 9\)</span></p>
<p><span class="math">\(H \times F = BI \Rightarrow 3 \times 9 = BI \Rightarrow B = 2, I = 7\)</span></p>
<p><span class="math">\(H \times B = D \Rightarrow 3 \times 2 = D \Rightarrow D = 6\)</span></p>
<p><span class="math">\(H \times I = BG \Rightarrow 3 \times 7 = 2G \Rightarrow G = 1\)</span></p>
<p><span class="math">\(H \times D = GA \Rightarrow 3 \times 6 = 1A \Rightarrow A = 8\)</span></p>
<p><span class="math">\(H \times A = BC \Rightarrow 3 \times 8 = 2C \Rightarrow C = 4\)</span></p>
<p>... and before we know it, it is solved!</p>
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