Weekly Challenge 50: How Old?
Let Joseph be $x$ years old now, Jasmine be $y$ years old now, and
me be $z$ years old now.
From the given information, we find three simulataneous
equations.
$x+1=2(y+1) \Rightarrow x=2y+2$
$z+6=k(x+6+y+6) \Rightarrow z=k(3y+13)-6$
$z+m=66=x+m+y+m \Rightarrow z=3y+1+m$
where k and m are some positive integers.
From this we deduce: $$66=3y+1+2m \Rightarrow m=\frac{65-3y}{2}
$$
We know the two equations: $z=66-m=66-\frac{65-3y}{2}$ and
$z=k(3y+13)-6$
Equating the two, we find: \begin{eqnarray*}
66-\frac{65-3y}{2} &=& k(3y+13)-6 \\ 27 &=&
y(6k-3) \end{eqnarray*}
If $k< 3$ then y is not an integer. So $k=2 \Rightarrow
y=3$
From this it is easy to find that x=7 and z=38.