# The GCHQ Christmas challenge

# Introduction

A few weeks ago I came across an article on the BBC news web site. It describes the Christmas challenge set by the UK Government Communications Headquarters (GCHQ) during the run up to Christmas 2023. The challenge was for school children aged 11-18 years.

The complete details can be found on the GCHQ’s challenge page.

The fourth puzzle (of seven) caught my eye, since it involves numbers.
So I thought I’ll have a go, even though I don’t fit the age range
`;-)`

# The challenge

Here is the fourth puzzle:

\[MI \times MI = MAA\] \[TI + TI = RA\] \[DO - SO + TI - MI = RE\] \[RE \times RE = \, ??\]Where, *Each letter represents a different digit*. We are asked to
find a *one-word answer which can follow Christmas*.

# An example solution

What follows is the tidied up version of my scribbles on the whiteboard. Of course, there are other ways of solving the puzzle.

From the second equation we can see that \(2 \times TI = RA\), so \(RA\) is an even number, and hence \(A \in \{ 0,2,4,6,8 \}\)

The third equation can be rewritten as follows:

\[(10D + O) - (10S + O) + (10T + I) - (10M + I) = RE\]rearranging the above, with each of the pairs of \(O\)s and \(I\)s cancelling themselves, we end up with the following equation:

\[10(D - S + T - M) = RE\]so, \(E=0\), and hence \(A\) is reduced to the set \(\{2,4,6,8\}\).

Now, back to the first equation, the one that I could have started with!

\[(MI)^2 = MAA\]So, \(MAA\) is a 3-digit square number with repeated digits on the right. This restricts the \(MI\) to the range \([10 .. 31]\), so we have four possibilities: 100, 144, 400 and 900.

We can exclude \(100\), \(400\) and \(900\), since they would imply that \(A=0\), but \(0\) already belongs to \(E\).

This leaves us with \(MI=12\) and \(MAA=144\).

We now have \(E=0\), \(M=1\), \(I=2\) and \(A=4\).

The second equation can now be written as \(T2+T2=R4\), so \(2T=R\), making \(R\) an even number, and hence \(R\in\{6,8\}\). This leads to \(T\in\{3,4\}\), but \(4\) belongs to \(A\), and so \(T=3\), and \(R=6\).

Now, \(RE=60\), which leads to the answer as follows:

\[RE \times RE = 60 \times 60 = 3600 = TREE\]So, **TREE** is the *one-word answer which can follow Christmas*.

# Carrying on …

The puzzle is solved, but I’m on a roll, so let’s solve for the rest of the letters.

So far we have \(E=0\), \(M=1\), \(I=2\), \(T=3\), \(A=4\) and \(R=6\).

From the simplified version of the third equation:

\[10(D - S + T - M) = RE\] \[D - S + 3 - 1 = 6\] \[D - S = 4\]From the list of unassigned digits, \(D,S \in \{ 5, 7, 8, 9 \}\), and the only pair of digits satisfying the above equation are \(5\) and \(9\), i.e.

\[D=9, S=5\]But, what about \(O\)? Well, we can start with \(O\in\{7,8\}\). However, since \(O\) is only mentioned in the third equation, in a self-cancelling manner, and with only 9 letters incorporated in the puzzle, we cannot narrow it down any further!

Have a successful and enjoyable 2024 :-)