15. (15 points) A cube with an edge length of 6 is cut into several smaller cubes with integer edge lengths. If the total surface area of these smaller cubes is $\frac{10}{3}$ times the surface area of the original large cube, find the number of smaller cubes with an edge length of 1.
15. (15 points) A cube with an edge length of 6 is cut into several smaller cubes with integer edge lengths. If the total surface area of these smaller cubes is $\frac{10}{3}$ times the surface area of the original large cube, find the number of smaller cubes with an edge length of 1.
【Solution】Solution: The surface area of the large cube: $6 \times 6 \times 6=216$, the volume is: $6 \times 6 \times 6=216$. The total surface area of the smaller cubes after cutting is: $216 \times \frac{10}{3}=720$.
Assume there is 1 smaller cube with an edge length of 5, then the remaining smaller cubes can only have an edge length of 1, and the number of these is: $\left(6^{3}-\right.$ $5^{3}$ ) $\div 1^{3}=91$ (units). At this time, the total surface area is: $5^{2} \times 6+1^{2} \times 6 \times 91=696 \neq 720$, so it is impossible to have a smaller cube with an edge length of 5.
(1) Similarly, the smaller cube with an edge length of 4 can be at most 1. In this case, it is impossible to have a smaller cube with an edge length of 3, and the remaining can only be cut into smaller cubes with an edge length of 2 or 1. Let the number of smaller cubes with an edge length of 2 be $a$, and the number of smaller cubes with an edge length of 1 be $b$,
$$
\text { then }\left\{\begin{array}{l}
2^{3} a+1^{3} b+4^{3}=216 \\
2^{2} \times 6 a+1^{2} \times 6 b+4^{2} \times 6=720
\end{array}\right.
$$

Solving, we get: $\left\{\begin{array}{l}a=12 \\ b=56\end{array}\right.$
(2) The number of smaller cubes with an edge length of 3 must be less than $(6 \div 3) \times(6 \div 3) \times(6 \div 3)=8$. Let the number of smaller cubes with an edge length of 2 be $a$, the number of smaller cubes with an edge length of 1 be $b$, and the number of smaller cubes with an edge length of 3 be $c$,
$$
\left\{\begin{array}{l}
3^{3} c+2^{3} a+1^{3} b=216 \\
3^{2} \times 6 c+2^{2} \times 6 a+1^{2} \times 6 b=720
\end{array}\right.
$$

Simplifying: $\left\{\begin{array}{l}27 c+8 a+b=216 \\ 9 c+4 a+b=120\end{array}\right.$
From the above, we get:
$b=9 c+24, a=\frac{120-9 c-b}{4}$,
When $c=0$, $b=24, a=24$,
When $c=1$, $b=33, a=19.5$, (not valid, discard)
When $c=2$, $b=42, a=15$,
When $c=3$, $b=51, a=10.5$, (not valid, discard)
When $c=4$, $b=60, a=6$,
When $c=5$, $b=69, a=28.5$, (not valid, discard)
When $c=6$, $b=78, a=-3$, (not valid, discard)
When $c=7$, $a=$ negative, (not valid, discard)
Therefore, the number of smaller cubes with an edge length of 1 can only be: 56 or 24 or 42 or 60.
Answer: The number of smaller cubes with an edge length of 1 can only be: 56 or 24 or 42 or 60.