$a=2$, $f=8$, $l=7$, $w=6$ and $y=5$.
We note first that $y=5$ since that is the only non-zero digit
that, when it is multiplied by 3, has itself as the units digit. So
there is a carry of 1 into the tens column. We note also that $a=1$
or $a=2$ as "$fly$"$< 1000$ and therefore $3\times$ "$fly$"$<
3000$. We now need $3\times l+1$ to end in either 1 or 2 and the
only possibility is $l=7$, giving $a=2$ with a carry of 2 into the
hundreds column. As $a=2$, $f$ must be at least 6. However, if
$f=6$ then $w=0$ which is not allowed. Also, the letters represent
different digits, so $f\neq 7$ and we can also deduce that $f\neq9$
since $f=9$ would make $w=9$. Hence $f=8$, making $w=6$ and the
letters represent $875\times 3=2625$.
This problem is taken from the UKMT Mathematical Challenges.