Let $N$ be the smallest number that has 378 divisors and is of the form $2^{a} \times 3^{b} \times 5^{c} \times 7^{d}$. What is the value of each of these exponents?
Since 2, 3, 5, and 7 are primes, for $2^{a} \times 3^{b} \times 5^{c} \times 7^{d}$ to have 378 divisors, we must have:

$$
(a+1) \times(b+1) \times(c+1) \times(d+1)=378
$$

Decomposing 378 into prime factors, we find $378=2 \times 3^{3} \times 7$. Therefore,

$$
(a+1) \times(b+1) \times(c+1) \times(d+1)=2 \times 3^{3} \times 7
$$

On the other hand, since $N$ is minimal, the exponents are ordered from the largest to the smallest, i.e., $a \geq b \geq c \geq d$.

We claim that $d>0$, because if $d=0$ then $a+1, b+1$ or $c+1$ has two factors greater than 1. If $a+1=m n$ with $m \geq n>1$, we have

$$
2^{a}=2^{m n-1}=2^{m-1} 2^{m n-m}=2^{m-1}\left(2^{m}\right)^{n-1} \geq 2^{m-1} 8^{n-1}>2^{m-1} 7^{n-1}
$$

where in the penultimate inequality we use the fact that $m \geq 3$. Thus, $2^{a} 3^{b} 5^{c} 7^{d}>2^{m-1} 3^{b} 5^{c} 7^{n-1}$, so we find a number with the same number of divisors but smaller. The proof is the same in the case where $b+1$ has two factors or $c+1$ has two factors. Therefore, $d \geq 1$ and we have the following unique possibilities:

| $a$ | $b$ | $c$ | $d$ | $(a+1)(b+1)(c+1)(d+1)=378$ |
| :---: | :---: | :---: | :---: | :---: |
| 20 | 2 | 2 | 1 | $21 \times 3 \times 3 \times 2$ |
| 13 | 2 | 2 | 2 | $14 \times 3 \times 3 \times 3$ |
| 8 | 6 | 2 | 1 | $9 \times 7 \times 3 \times 2$ |
| 6 | 5 | 2 | 2 | $7 \times 6 \times 3 \times 3$ |

Finally, since

$$
\begin{aligned}
& \frac{2^{20} \cdot 3^{2} \cdot 5^{2} \cdot 7^{1}}{2^{13} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2}}=\frac{2^{7}}{7}>1 \\
& \frac{2^{13} \cdot 3^{2} \cdot 5^{2} \cdot 7^{2}}{2^{8} \cdot 3^{6} \cdot 5^{1} \cdot 7^{1}}=\frac{2^{5} \cdot 7}{3^{4}}>1
\end{aligned}
$$

and

$$
\frac{2^{8} \cdot 3^{6} \cdot 5^{2} \cdot 7^{1}}{2^{6} \cdot 3^{5} \cdot 5^{2} \cdot 7^{2}}=\frac{2^{2} \cdot 3}{7}>1
$$

we conclude that the value of $N$ is $2^{6} \cdot 3^{5} \cdot 5^{2} \cdot 7^{2}$. Therefore, $a=6, b=5, c=2$, and $d=2$.