I pick two whole numbers x and y between 1 and 10 inclusive (not necessarily distinct). My friend picks two numbers x -4 and 2y-1. If the pr

We can write the equation:

 

(x-4) * (2y-1) = 1 +x*y 

which is what the problem gives us

It then asks us to find xy

Let's expand out the left side first. We get:

 

2xy-x-8y +4 = 1 + xy

Grouping everything on one side, we get:

 

xy - x - 8y + 3 = 0

Now, we can make handy of a principle known (on aops' forum at least) as "simon's favorite factoring trick", or "completing the rectangle".

 

if we add 5 on both sides, we can factor the left hand side, into:

 

(x - 8 ) * (y - 1) = 5

The problem then becomes a lot simpler, since 5 has two integer factors: 5 and 1

Let's assume x - 8 = 5 and y - 1 = 1

Then we have:

x = 13

y = 2

Clearly,  13>10, so this solution is not correct.

Let's assume the other case:

where,

x - 8 = 1

and

y - 1 = 5

Then we have:

x = 9

y = 6

Clearly, this fits our criterion, giving us the answers of 

9 * 6 = 54