### Instruction:
Find the area of the region described by $x \ge 0,$ $y \ge 0,$ and
\[100 \{x\} \ge \lfloor x \rfloor + \lfloor y \rfloor.\]Note: For a real number $x,$ $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x.$  For example, $\{2.7\} = 0.7.$

### Response:

Let $n = \lfloor x \rfloor,$ and let $\{x\} = (0.x_1 x_2 x_3 x_4 \dots)_{10},$ so the $x_i$ are the decimal digits.  Then the given condition becomes
\[\lfloor y \rfloor \le 100 \{x\} - \lfloor x \rfloor = (x_1 x_2.x_3 x_4 \dots)_{10} - n.\]Since $\lfloor y \rfloor$ is an integer, this is equivalent to
\[\lfloor y \rfloor \le (x_1 x_2)_{10} - n.\]First, let's look at the interval where $0 \le x < 1,$ so $n = 0.$  For $0 \le x < 0.01,$ we want
\[\lfloor y \rfloor \le 0,\]so $0 \le y < 1.$

For $0.01 \le x < 0.02,$ we want
\[\lfloor y \rfloor \le 1,\]so $0 \le y < 2.$

For $0.02 \le x < 0.03,$ we want
\[\lfloor y \rfloor \le 2,\]so $0 \le y < 3,$ and so on.

Thus, for $0 \le x < 1,$ the region is as follows.

[asy]
unitsize(1 cm);

draw((0,0)--(6,0));
draw((0,0)--(0,6));
filldraw((0,0)--(0,1)--(1,1)--(1,0)--cycle,gray(0.7));
filldraw((1,0)--(1,2)--(2,2)--(2,0)--cycle,gray(0.7));
filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle,gray(0.7));
filldraw((5,0)--(5,6)--(6,6)--(6,0)--cycle,gray(0.7));

label("$0$", (0,0), S, fontsize(10));
label("$0.01$", (1,0), S, fontsize(10));
label("$0.02$", (2,0), S, fontsize(10));
label("$0.03$", (3,0), S, fontsize(10));
label("$0.99$", (5,0), S, fontsize(10));
label("$1$", (6,0), S, fontsize(10));
label("$0$", (0,0), W, fontsize(10));
label("$1$", (0,1), W, fontsize(10));
label("$2$", (0,2), W, fontsize(10));
label("$3$", (0,3), W, fontsize(10));
label("$100$", (0,6), W, fontsize(10));
label("$\dots$", (4,2));
label("$\vdots$", (0,4.5), W);
[/asy]

The area of this part of the region is then
\[0.01(1 + 2 + 3 + \dots + 100) = 0.01 \cdot \frac{100 \cdot 101}{2}.\]Next, we look at the interval where $1 \le x < 2,$ so $n = 1.$  For $1 \le x < 1.01,$ we want
\[\lfloor y \rfloor \le 0 - 1 = -1,\]so there are no values of $y$ that work.

For $1.01 \le x < 1.02,$ we want
\[\lfloor y \rfloor \le 1 - 1 = 0,\]so $0 \le y < 1.$

For $1.02 \le x < 1.03,$ we want
\[\lfloor y \rfloor \le 2 - 1 = 1,\]so $0 \le y < 2,$ and so on.

Thus, for $1 \le x < 2,$ the region is as follows.

[asy]
unitsize(1 cm);

draw((0,0)--(6,0));
draw((0,0)--(0,5));
filldraw((1,0)--(1,1)--(2,1)--(2,0)--cycle,gray(0.7));
filldraw((2,0)--(2,2)--(3,2)--(3,0)--cycle,gray(0.7));
filldraw((5,0)--(5,5)--(6,5)--(6,0)--cycle,gray(0.7));

label("$1$", (0,0), S, fontsize(10));
label("$1.01$", (1,0), S, fontsize(10));
label("$1.02$", (2,0), S, fontsize(10));
label("$1.03$", (3,0), S, fontsize(10));
label("$1.99$", (5,0), S, fontsize(10));
label("$2$", (6,0), S, fontsize(10));
label("$0$", (0,0), W, fontsize(10));
label("$1$", (0,1), W, fontsize(10));
label("$2$", (0,2), W, fontsize(10));
label("$3$", (0,3), W, fontsize(10));
label("$99$", (0,5), W, fontsize(10));
label("$\dots$", (4,2));
label("$\vdots$", (0,4), W);
[/asy]

The area of this part of the region is then
\[0.01(1 + 2 + 3 + \dots + 99) = 0.01 \cdot \frac{99 \cdot 100}{2}.\]Similarly, the area of the region for $2 \le x < 3$ is
\[0.01(1 + 2 + 3 + \dots + 98) = 0.01 \cdot \frac{98 \cdot 99}{2},\]the area of the region for $3 \le x < 4$ is
\[0.01(1 + 2 + 3 + \dots + 97) = 0.01 \cdot \frac{97 \cdot 98}{2},\]and so on, until the area of the region for $99 \le x < 100$ is
\[0.01(1) = 0.01 \cdot \frac{1 \cdot 2}{2}.\]Hence, the total area of the region is
\[\frac{0.01}{2} (1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + 100 \cdot 101) = \frac{1}{200} \sum_{k = 1}^{100} k(k + 1).\]To compute this sum, we can use the formula
\[\sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}.\]Alternatively, we can write
\[k(k + 1) = \frac{(k + 2) - (k - 1)}{3} \cdot k(k + 1) = \frac{k(k + 1)(k + 2) - (k - 1)k(k + 1)}{3},\]which allows sum to telescope, and we get
\[\frac{1}{200} \sum_{k = 1}^{100} k(k + 1) = \frac{1}{200} \cdot \frac{100 \cdot 101 \cdot 102}{3} = 1717.\]