Putnam (2021): Solution to problem A1

Since each hop is of length $5$ and only integral positions are allowed, that means the grasshopper can move in any of the following ways:

• $(0, \pm 5)$ or $(\pm 5, 0)$
• $(\pm 3, \pm 4)$ or $(\pm 4, \pm 3)$

The first of these is straightforward; the second group is because $(3, 4, 5)$ are the side lengths of a right triangle.

To go from $(0, 0)$ to $(2021, 2021)$ we need to move as efficiently as possible diagonally across both $x$ and $y$ directions, so taking alternating $(3, 4)$ and $(4, 3)$ steps moves the grasshopper closer to the goal by $(7, 7)$ every two steps.

Note: the alternative of using steps $(5, 0)$ and $(0, 5)$ in either order only moves us by $(5, 5)$ every two steps instead, which is not nearly as efficient.

Unfortunately, $2021$ is not divisible by $7$, so the grasshopper will need to make some adjustments as it gets closer to the target.

Since $\lfloor 2021 / 7 \rfloor = 288$, the grasshopper initially will need to make $288 \cdot 2 = \mathbf{576}$ hops (since the grasshopper needs $2$ hops to move $(7, 7)$ diagonally), and the grasshopper will actually end up at $(288 \cdot 7, 288 \cdot 7) = (2016, 2016)$.

This puts the grasshopper $(2021, 2021) - (2016, 2016) = (5, 5)$ away from its goal, which it can reach in $\mathbf{2}$ additional hops: $(5, 0)$ and $(0, 5)$ in either order.

Thus, the total number of hops the grasshopper will need is $576 + 2 = \boxed{578}$ hops.