# British Maths Olympiad (BMO) 2005 Round 1 Question 1 how to make progress?

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.

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.

• Note : I would like you to see if you can eliminate $y$ instead of $x$ now, and get a relation between $n$ and $x$ that you can exploit. In general, one could not have observed the equations given to you and decided which variable to eliminate, and further the temptation is to start by eliminating $n$ since it is the subject of the question. In this respect, I offer (much advanced) sympathy. – астон вілла олоф мэллбэрг Apr 29 '18 at 8:46

First note that we have some lower bounds: Not only are $x,y,n$ positive, but the amounts $y-3$ and $x-n$ must also be positive. In fact, if $y=4$, we get $x=n-3$, contradicting $x>n$; we conclude that $y\ge 5$

Plugging the value of $x$ from the first into the second equation, we get $$\tag1y+n=3(n(y-3)-3-n)$$ which we can rearrange into the following equivalent equations, where we first expand and bring all to one side and then play around to find a product of linear factors that resembles the total as closely as poyyible: $$\tag23yn-13n-y-9=0.$$ $$\tag3(y-4)(3n-1)=n+13$$ $$\tag4(y-5)(3n-1)=14-2n.$$ Especially $(4)$ raises our interest: We already know that $y\ge 5$. Hence the left hand side is non-negative and we must have $14-2n\ge 0$, i.., $n\in\{1,2,3,4,5,6,7\}$. For each of these cases, we can compute $y=\frac{14-2n}{3n-1}+5$ and then $x=n(y-3)-3$: $$\begin{matrix} n=1,&y=11,&x=5\\ n=2,&y=7,&x=5\\ n=3,&y=6,&x=6\\ n=4,&y\notin \Bbb N\\ n=5,&y\notin \Bbb N\\ n=6,&y\notin \Bbb N\\ n=7,&y=5,&x=11\\ \end{matrix}$$ among which we detect precisely four solutions.