The squares of a chessboard are labelled with numbers, as shown below.

[asy]
unitsize(0.8 cm);

int i, j;

for (i = 0; i <= 8; ++i) {
  draw((i,0)--(i,8));
  draw((0,i)--(8,i));
}

for (i = 0; i <= 7; ++i) {
for (j = 0; j <= 7; ++j) {
  label("$\frac{1}{" + string(i + 8 - j) + "}$", (i + 0.5, j + 0.5));
}}
[/asy]

Eight of the squares are chosen, so that there is exactly one chosen square in each row and each column.  Find the minimum sum of the labels of the eight chosen squares.
Numbers the rows 1, 2, 3, $\dots,$ 8 from top to bottom.  Let $r_1$ be the row number of the chosen square in the first column.  (For example, if the 5th square is chosen in the first column, then $r_1 = 5.$)  Then the label of that square is $\frac{1}{r_1}.$

Similarly, if $r_2$ is the row number of the chosen square in the second column, then its label is
\[\frac{1}{r_2 + 1}.\]In general, let $r_i$ be the row number of the chosen square in column $i,$ so its label is
\[\frac{1}{r_i + i - 1}.\]Then we want to minimize
\[\frac{1}{r_1} + \frac{1}{r_2 + 1} + \frac{1}{r_3 + 2} + \dots + \frac{1}{r_8 + 7}.\]By AM-HM,
\[\frac{r_1 + (r_2 + 1) + (r_3 + 2) + \dots + (r_8 + 7)}{8} \ge \frac{8}{\frac{1}{r_1} + \frac{1}{r_2 + 1} + \frac{1}{r_3 + 2} + \dots + \frac{1}{r_8 + 7}},\]so
\begin{align*}
\frac{1}{r_1} + \frac{1}{r_2 + 1} + \frac{1}{r_3 + 2} + \dots + \frac{1}{r_8 + 7} &\ge \frac{64}{r_1 + (r_2 + 1) + (r_3 + 2) + \dots + (r_8 + 7)} \\
&= \frac{64}{r_1 + r_2 + r_3 + \dots + r_8 + 28}.
\end{align*}Since there exists one chosen square in each row, $r_1,$ $r_2,$ $r_3,$ $\dots,$ $r_8$ are equal to 1, 2, 3, $\dots,$ 8 in some order.  Therefore,
\[\frac{1}{r_1} + \frac{1}{r_2 + 1} + \frac{1}{r_3 + 2} + \dots + \frac{1}{r_8 + 7} \ge \frac{64}{1 + 2 + 3 + \dots + 8 + 28} = \frac{64}{36 + 28} =  1.\]Equality occurs when we choose all eight squares labelled $\frac{1}{8},$ so the smallest possible sum is $\boxed{1}.$
1
Level 3
Intermediate Algebra
Problem: The squares of a chessboard are labelled with numbers, as shown below.

[asy]
unitsize(0.8 cm);

int i, j;

for (i = 0; i <= 8; ++i) {
  draw((i,0)--(i,8));
  draw((0,i)--(8,i));
}

for (i = 0; i <= 7; ++i) {
for (j = 0; j <= 7; ++j) {
  label("$\frac{1}{" + string(i + 8 - j) + "}$", (i + 0.5, j + 0.5));
}}
[/asy]

Eight of the squares are chosen, so that there is exactly one chosen square in each row and each column.  Find the minimum sum of the labels of the eight chosen squares.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum sum of the labels of the eight chosen squares on the chessboard, we need to analyze the structure of the labels. Each square is labeled with the fraction $\frac{1}{i + 8 - j}$, where $i$ represents the column index and $j$ represents the row index. 

The goal is to select one square from each row and each column, which corresponds to selecting a permutation of the indices from 0 to 7. The sum of the labels for a specific selection can be expressed as:

Sum = $\sum_{k=0}^{7} \frac{1}{i_k + 8 - j_k}$

where $(i_k, j_k)$ are the indices of the chosen squares. 

To minimize this sum, we need to maximize the denominators $i_k + 8 - j_k$. The maximum value of $j_k$ is 7 (the last row), which means we want to choose squares such that $j_k$ is as large as possible for the corresponding $i_k$. 

The optimal arrangement occurs when we select squares from the main diagonal of the chessboard, where $i_k = j_k$. This gives us:

Sum = $\sum_{k=0}^{7} \frac{1}{k + 8 - k} = \sum_{k=0}^{7} \frac{1}{8} = 8 \cdot \frac{1}{8} = 1$.

Thus, the minimum sum of the labels of the eight chosen squares is achieved when each square corresponds to the indices where $i_k = j_k$, leading to the final answer of 1.<|END_OF_LATENT|> The final answer is $1$.