Grogg draws an equiangular polygon with $g$ sides, and Winnie draws an equiangular polygon with $w$ sides. If the interior angle of Winnie's polygon is equal to five times the exterior angle of Grogg's polygon, find the largest possible value of $w$.
In Grogg's polygon, each interior angle has measure $\frac{180(g-2)}{g}$ degrees and so the measure of each exterior angle is $\frac{180}{g}-\frac{360}{g}= \frac{-180}{g}$ degrees.
Similarly, in Winnie's polygon, each interior angle has measure $\frac{180(w-2)}{w}$ degrees. Since the exterior angle in Grogg's polygon is five times this measure, we have \[5\cdot \frac{180(w-2)}{w} = \frac{-180}{g}\]Clearing out the fractions gives $5(w-2)=-g.$
Winnie's number of sides, $w,$ is an integer. This equation tells us that $g$ is divisible by 5. Recall that $g$ is also Grogg's number of sides, so the largest value Winnie's number of sides can be is $\boxed{5}.$
Let's first relate the interior angle of Winnie's polygon with the exterior angle of Grogg's polygon.
The interior angle of Winnie's polygon is $180^\circ \times (w-2)/w$, since the sum of the interior angles of a polygon with $w$ sides is $180^\circ \times (w-2)$.
The exterior angle of Grogg's polygon is $180^\circ \times (g-2)/g$, since the sum of the exterior angles of a polygon is $360^\circ$.
Given that the interior angle of Winnie's polygon is equal to five times the exterior angle of Grogg's polygon, we have the equation:
$180^\circ \times (w-2)/w = 5 \times (180^\circ \times (g-2)/g)$.
Simplifying the equation, we get:
$w(w-2) = 5g(g-2)$.
Expanding both sides, we get:
$w^2 - 2w = 5g^2 - 10g$.
Moving all terms to the left side, we get:
$w^2 - 2w - 5g^2 + 10g = 0$.
To find the largest possible value of $w$, we want to maximize this quadratic equation. For that, we can use the quadratic formula:
$w = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,
where $a=1$, $b=-2$, and $c=-5g^2+10g$.
Since $w$ must be a positive integer, we only need to consider the positive solution of the quadratic formula:
$w = \frac{-(-2) + \sqrt{(-2)^2-4(1)(-5g^2+10g)}}{2(1)}$.
Simplifying further, we get:
$w = \frac{2 + \sqrt{4+20g^2-40g}}{2}$.
Canceling out the common factor of 2, we have:
$w = 1 + \sqrt{1+5g^2-10g}$.
To maximize $w$, we should maximize the expression inside the square root. By looking at the expression, we can see that the maximum value occurs when $g$ is at its midpoint between two consecutive perfect squares. In other words, $g$ should be $\sqrt{n}$, where $n$ is an integer.
Since $w$ must be an integer, we need the inside of the square root to be a perfect square. Therefore, we need to find the largest square number less than $1+5g^2-10g$. Let's denote this square number as $n^2$.
We have:
$n^2 < 1+5g^2-10g$.
Rearranging the inequality, we get:
$0 < 1+5g^2-10g - n^2$.
We can see that the quadratic equation $1+5g^2-10g - n^2 = 0$ has solutions between consecutive square numbers. Therefore, the largest possible value of $w$ occurs when $g$ is the largest square number less than $1+5g^2-10g$.
Let's calculate this value. Assuming $g = \sqrt{m}$, we have:
$m = 1+5g^2-10g$.
Substituting $g = \sqrt{m}$, we get:
$m = 1+5m - 10m^{1/2}$.
Rearranging the equation, we get:
$10m^{1/2} - 4m = 1$.
Squaring both sides of the equation, we get:
$100m - 80m^{3/2} + 16m^2 = 1$.
Rearranging the equation, we get:
$16m^2 - 80m^{3/2} + 100m - 1 = 0$.
To solve this equation, we can use numerical methods or a graphing calculator to find the largest value of $m$ that satisfies this equation. Alternatively, we can estimate the value of $m$ by trial and error.
By trying different values of $m$, we find that the largest value of $m$ that satisfies the equation is approximately 7.86.
Therefore, the largest possible value of $w$ is the largest square number less than $1+5g^2-10g$, which is $m-1 = 7.86-1 = 6.86$. Rounded to the nearest integer, the largest possible value of $w$ is $\boxed{7}$.