$35 \cdot 1220$ unit cubes are stacked as shown in the figure. First, the first layer consists of 10 cubes arranged in a triangular pattern; then the second layer is 6 cubes arranged in a triangular pattern placed in the center of the first layer; the third layer is 3 cubes arranged in a triangular pattern placed in the center of the second layer; and finally, 1 cube is placed in the center of the third layer. The 10 cubes in the bottom layer are labeled with numbers from 1 to 10 in some order, and the number on each cube in the second, third, and fourth layers is the sum of the numbers on the 3 cubes it rests on.
The smallest possible value of the number on the top cube is
(A) 55 .
(B) 83 .
(C) 114 .
(D) 137 .
(E) 144 .
(44th American High School Mathematics Examination, 1993)
[Solution] Note that the center cell of the first layer is calculated the most times, so it should be marked as 1, and those cells at the corners are used the least, so they should be marked as $8,9,10$. For example, to get the correct answer, mark the cells of the bottom layer as shown in the diagram.
$$

$$

More generally, let the cells of the bottom layer be marked as
$$
\begin{array}{l}
v_{1} \\
\begin{array}{ll}
e_{1} & e_{2}
\end{array} \\
\begin{array}{lll}
e_{3} & c & e_{4}
\end{array} \\
\begin{array}{llll}
v_{2} & e_{5} & e_{6} & v_{3}
\end{array} \\
\end{array}
$$

Then the cells of the second layer are marked as
$$
\begin{array}{c}
\left(e_{1}+e_{2}+v_{1}\right) \\
\left(c+e_{1}+e_{3}\right) \quad\left(c+e_{2}+e_{4}\right) \\
\left(e_{3}+e_{5}+v_{2}\right) \quad\left(c+e_{5}+e_{6}\right) \quad\left(e_{4}+e_{6}+v_{3}\right)
\end{array}
$$

The cells of the third layer are marked as
$$
\begin{array}{c}
\left(2 c+2 e_{1}+2 e_{2}+e_{3}+e_{4}+v_{1}\right) \\
\left(2 c+e_{1}+2 e_{3}+2 e_{5}+e_{6}+v_{2}\right) \quad\left(2 c+e_{2}+2 e_{4}+e_{5}+2 e_{6}+v_{3}\right)
\end{array}
$$

So $t=6 c+3\left(e_{1}+e_{2}+e_{3}+e_{4}+e_{5}+e_{6}\right)+\left(v_{1}+v_{2}+v_{3}\right)$ is the number marked on the top cell.

Therefore, when $c=1,\left\{e_{1}, e_{2}, \cdots, e_{6}\right\}=\{2,3, \cdots, 7\}$, and $\left\{v_{1}, v_{2}, v_{3}\right\}=\{8,9,10\}$, $t$ reaches its minimum value. So the minimum value of $t$ is
$$
6 \cdot 1+3 \cdot(2+3+\cdots+7)+(8+9+10)=114 .
$$

Thus, the answer is $(C)$.