Hamilton 2018 Q1

Comment
There are many approaches to this question. It is straightforward to show that, given a solution \(m = M\) and \(n = N\), the pair \(m = (M - 9k)\) and \(n = (N + 10k)\) is also a solution for any integer value of \(k\). This allows one to ‘spot’ a solution and generate a set of solutions from that one.

However, if we use this approach, it is important to show that there aren't any other solutions which are not in this set. (Indeed, in any question where we are trying to find all possible solutions, finding them is only half the battle − a key step is to show that we have found them all.)

The first of these solutions shows a way to complete the argument introduced above; the second demonstrates a slightly different approach.

Solution 1
Let \(M\) and \(N\) be a solution to the given equation, so \(20M + 18N = 2018\).

For any integer \(k\), the pair \(m = (M - 9k)\) and \(n = (N + 10k)\) is also a solution, since

It is easy to verify (and not too difficult to spot) that \(m = 100\), \(n = 1\) is a solution.

This quickly leads us to the following twelve solutions, which are all of the form \(m = 100 - 9k\), \(n = 1 + 10k\) for \(k = 0,\ 1,\ 2,\ \ldots,\ 11\):

We will now show that any solution to the given equation must be of the form \(m = 100 - 9k\), \(n = 1 + 10k\).

Let \((M, N)\) be a pair of integers for which \(20M + 18N = 2018\).

Since \(20 \times 100 + 18 \times 1 = 2018\), we can subtract to give

Then

Since 10 and 9 have no common factor greater than 1, \((100 - M)\) must be a multiple of 9, so we can write \(100 - M = 9k\) for some integer \(k\). This leads to \(M = 100 - 9k\).

Then we have, from \((\ast)\), \(90k = 9(N - 1)\), which leads to \(N = 1 + 10k\).

So we have now shown that any pair of integers \((M, N)\) which are a solution to the equation must be of the form \((100 - 9k,\ 1 + 10k)\) for some integer \(k\).

For \(100 - 9k\) to be positive we need \(k \leqslant 11\) and for \(1 + 10k\) to be positive we need \(k \geqslant 0\), so the twelve solutions listed above are the only twelve.

Solution 2
If \(20m + 18n = 2018\) then

Since \(m\) is an integer, \(9n + 1\) must be a multiple of 10, which means that \(9n\) must be one less than a multiple of 10. For this to be true, the final digit of \(9n\) must be 9, which happens exactly when the final digit of \(n\) is 1.

Every value of \(n\) with final digit 1 will produce a corresponding value of \(m\). No other value of \(n\) is possible (and hence no other possible values of \(m\) possible either).

So now we just need to count the number of positive values of \(n\) with final digit 1 which also give a positive value of \(m\).

Since \(m\) is positive, \(\dfrac{9n + 1}{10}\) can be at most 100, with equality when \(n = 111\).

Hence \(0 < n \leqslant 111\). There are 12 values of \(n\) in this range which have final digit 1, and so there are 12 possible values of \(m\).

[Note that we have not found any of the solutions to the given equation; we did not need to do this in order to answer the question.]