[Math] British Maths Olympiad (BMO) 2005 Round 1 Question 1 how to make progress

algebra-precalculuscontest-mathelementary-number-theory

The question is as follows:

Each of Paul and Jenny has a whole number of pounds.
He says to her: “If you give me £3, I will have $n$ times as much as you”.
She says to him: “If you give me £$n$, I will have 3 times as much
as you”.

Given that $n$ is a positive integer what are the possible values for $n$?

We begin by representing Paul and Jenny's amounts by $x,y$ respectively.

Then we have:

$x+3 = n(y-3)\\y+n = 3(x-n)$

After some manipulation we can arrive at:

$n = (x+3)/(y-3) = (3x-y)/(4)$

If we let $y-3 = 4$ and $x+3 = 3x-y$ we get a solution but beyond that I'm not quite sure how to progress.

General ideas on how to approach problems like this would be appreciated as well as some help with the solution.

Best Answer

I think you are trying to look only at how $n$ is expressed in terms of the rest of the variables. Indeed, such an approach is good, but you can do with other variables as well! So let's try to write $n$ in terms of $y$, by eliminating $x$ from both equations.

From the first, $x = ny-3n-3$. From the second, $x = \frac{y+n}{3} + n$.

Combining these, we get $\frac{y+n}{3} + n = ny-3n-3$,so multiplying by $3$, $y+4n = 3ny-9n-9$. Isolating $n$ gives $y + 9 = n(3y-13)$, and hence $$ n = \frac{y+9}{3y-13} $$

Note that $n$ is a positive integer. If $y > 12$, then $3y-13 > 22$ while $y + 9 < 22$, so that $n$ cannot possibly be an integer! Furthermore, $3y-13 < 0$ if $y < 5$, so we have $5 \leq y \leq 11$.

Now, we can test : $y = 5$ gives $n = 7$ and $x = 11$, which works out.

$y = 6$ gives $n = 3$ and $x = 6$, which works out.

$y = 7$ gives $n = 2$ and $x = 5$, which works out.

$y = 8,9,10$ don't work out. Finally, $y = 11$ gives $n = 1$ and $x = 5$ which works out.

Finally, $n = 1,2,3,7$.

Concluding, the difference of approach was in isolating a different variable from that of $n$. Remember that all the variables are related in a manner such that knowing one means you know the rest, so it was sufficient to deal with any of the variables. Finally, expressions involving $\frac{ay+b}{cy+d}$ are restrictive if integers, when $a,c$ are small.