TITLE: How many 1-balls are needed to cover the 2-ball in an $n$-dimensional Euclidean Space?
QUESTION [15 upvotes]: Consider $\mathbb{R}^n$ and its usual Euclidean norm given by the distance $d(x,y) = \sqrt{\sum_{i=1}^n (x_i-y_i)^2}$.
Let $B(y,1) = \{ x\in \mathbb{R}^n : d(x,y) \leq 1 \}$ be the closed 1-ball around $y$ and  $B(0,2) = \{x\in \mathbb{R}^n:  d(x,0) \leq 2 \}$ be the 2-ball centered at the origin. 
How many 1-balls do you need to cover (contain) the 2-ball?

REPLY [2 votes]: In $\mathbb{R}^2$ it's 7.
It's sufficient because:
You can take the circum-disks of the cells in a hexagonal grid as your small disks (http://mathworld.wolfram.com/HexagonalGrid.html). It's easy to see that a disk of radius 2, centered at the center of one hexagon, will intersect exactly 7 hexagons, and so the union of the circum-disks of those hexagons (which have radius 1) covers the disk of radius 2.
It's necessary, because: 
Any covering of the big disk must also cover its boundary. If you use the 6 disks as in the hexagonal arrangement, then they never overlap on the boundary, and the part of the arc covered by each one is maximal, this means that fewer than 6 disks won't be enough to cover the boundary. Then, since any disk that contains the origin will at most intersect the boundary in a single point, none of our 6 boundary disks can contain the origin, and thus an additional one is needed.