Tweedle Dum and Tweedle Dee

Let DUM begin with $x$ and DEE with $y$ coins. Note that $x < y$.
Following the giving of money algebraically the table shows DUM and DEE's fortunes after each gift:

DUM	DEE
$x$	$y$
$x+\frac{1}{3}y$	$\frac{2}{3}y$
$\frac{1}{2}x+\frac{1}{6}y$	$\frac{1}{2}x+\frac{5}{6}y$
$\frac{5}{6}x+\frac{13}{8}y$	$\frac{1}{6}x+\frac{5}{8}y$
$\frac{5}{24}x+\frac{13}{72}y$	$\frac{19}{24}x+\frac{59}{72}y$
$\frac{53}{72}x+\frac{157}{216}y$	$\frac{19}{72}x+\frac{59}{216}y$
$\frac{53}{108}x+\frac{157}{324}y$	$\frac{55}{108}x+\frac{167}{324}y$

After the last 'gift' of $1$ coin, we know that the totals are equal, so
\begin{align}
\frac{53}{108}x+\frac{157}{324}y+1 &= \frac{55}{108}x+\frac{167}{324}y -1 \\
2 &= \frac{2}{108}x+\frac{5}{162}7 \\
324 &= 3x + 5y
\end{align}
Integer solutions of this equation are found by noticing that since $3x$ and $5y$ appear, if we introduce $n$ such that $x = 5n+a$ and $y=-3n+b$ then $n$ will cancel from the equation so we may find values for $a$ and $b$. One such solution is
\begin{align}
x &= 5n+3 & y&=-3n +63,
\end{align}
but different values of $a$ and $b$ are possible. (Why? Can you convince yourself that as $n$ varies any correct values of $a$ and $b$ will give all the solutions?)
Since we know $0 < x < y$ we have the solutions:

$x$	$y$
$3$	$63$
$8$	$60$
$13$	$57$
$18$	$54$
$23$	$51$
$28$	$48$
$33$	$45$
$38$	$42$

Though you should still check that the gifts are still possible with these amounts - remember neither DUM nor DEE can ever give away exactly $1/3$ of a pound!