Problem 622

Suppose that the rectangle has width $w$ and length $\ell$, both of which are positive integers
with $w\leq \ell$.

The perimeter of the rectangle is $2(w+
$\ell)\$,$ and thus the perimeter is even (as a result of
multiplication by $2$).

The perimeter is also a multiple of $7$, and so the perimeter is an even
multiple of $7$.

The smallest even multiple of $7$ is
$14$.

If the perimeter is $14$, then $2(w+\ell)=14$ or $w+\ell=7$, and so the possible side
lengths, $(w,\ell)$, are $(1,6)$, $(2,5)$ and $(3,4)$.

The areas of the rectangles are $6\times1=6$, $2\times5=10$, and $3\times4$, respectively, none of which is
a multiple of $9$, and so $14$ is not the smallest possible
perimeter.

The next smallest even multiple of $7$ is $28$.

If the perimeter is $28$, then $2(w+\ell)=28$ or $w+\ell=14$, and so the possible side
lengths, $(w,\ell)$, are $(1,13)$, $(2,12)$, $(3,11)$, $(4,10)$, $(5,9)$, $(6,8)$, and $(7,7)$.

The areas of the rectangles are $13$,
$24$, $33$, $40$, $45$, $48$, and $49$, respectively, and $45$ is a multiple of $9$.

The rectangle with width $5$ and
length $9$ has an area that is a
multiple of $9$, a perimeter that is
a multiple of $28$, and so the
smallest possible perimeter is $28$.

Note that we could also approach this problem by first finding the side
lengths for which the area is a multiple of $9$, and then determining which of these
gives the smallest perimeter that is a multiple of $7$.