Solution:
Let $x$ be a real number. For any integer $n \geq 0$, $[2^n x]$ is the greatest integer less than or equal to $2^n x$.

Let $S(x) = [x] + [2x] + [4x] + [8x] + [16x] + [32x]$.

Let $x = a + y$, where $a$ is an integer and $0 \leq y < 1$.
Then $[2^k x] = [2^k a + 2^k y] = 2^k a + [2^k y]$ for $k = 0,1,2,3,4,5$.

So,
$$S(x) = \sum_{k=0}^5 [2^k x] = \sum_{k=0}^5 (2^k a + [2^k y]) = (\sum_{k=0}^5 2^k)a + \sum_{k=0}^5 [2^k y]$$
But $\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32 = 63$.
So,
$$S(x) = 63a + \sum_{k=0}^5 [2^k y]$$

Now, $0 \leq y < 1$, so $0 \leq 2^k y < 2^k$.
Therefore, $[2^k y]$ is an integer between $0$ and $2^k - 1$.

Thus, $\sum_{k=0}^5 [2^k y]$ is an integer between $0$ and $1 + 3 + 7 + 15 + 31 = 57$ (since $[2^k y] \leq 2^k - 1$), but actually, for $k=0$ to $5$, $[2^k y]$ runs from $0$ to $2^k - 1$, so the sum runs from $0$ to $1 + 3 + 7 + 15 + 31 = 57$ for $k=1$ to $5$, plus $[y]$ which is $0$ (since $0 \leq y < 1$), so the sum is from $0$ to $62$.

But let's check:
For $k=0$, $[2^0 y] = [y] = 0$ (since $0 \leq y < 1$).
For $k=1$, $[2y]$ runs from $0$ to $1$.
For $k=2$, $[4y]$ runs from $0$ to $3$.
For $k=3$, $[8y]$ runs from $0$ to $7$.
For $k=4$, $[16y]$ runs from $0$ to $15$.
For $k=5$, $[32y]$ runs from $0$ to $31$.
So the sum $\sum_{k=0}^5 [2^k y]$ runs from $0$ to $0 + 1 + 3 + 7 + 15 + 31 = 57$.

Therefore, $S(x) = 63a + m$, where $m$ is an integer between $0$ and $57$.

Thus, $S(x)$ can only take values congruent to $m$ modulo $63$, i.e., $S(x) \equiv m \pmod{63}$, where $0 \leq m \leq 57$.

But for a fixed $a$, $S(x)$ runs through $63a, 63a+1, \ldots, 63a+57$ as $y$ varies from $0$ to $1$.

Therefore, $S(x)$ can only take values congruent to $0,1,2,\ldots,57$ modulo $63$.

Now, $12345 = 63 \times 196 + 57$.
So $12345 \equiv 57 \pmod{63}$.

Therefore, for $a = 196$, $S(x)$ could be $63 \times 196 + m$ for $0 \leq m \leq 57$.
But $12345 = 63 \times 196 + 57$, so $m = 57$.

But let's check if $m = 57$ is possible.

But $m = \sum_{k=0}^5 [2^k y] = 0 + 1 + 3 + 7 + 15 + 31 = 57$.
Is it possible for $[2y]=1$, $[4y]=3$, $[8y]=7$, $[16y]=15$, $[32y]=31$ and $[y]=0$?

Let's see:
$[2y]=1$ implies $1 \leq 2y < 2$, so $0.5 \leq y < 1$.
$[4y]=3$ implies $3 \leq 4y < 4$, so $0.75 \leq y < 1$.
$[8y]=7$ implies $7 \leq 8y < 8$, so $0.875 \leq y < 1$.
$[16y]=15$ implies $15 \leq 16y < 16$, so $0.9375 \leq y < 1$.
$[32y]=31$ implies $31 \leq 32y < 32$, so $0.96875 \leq y < 1$.

So, $y$ must satisfy all these inequalities:
$0.96875 \leq y < 1$.

But $[y]=0$ requires $0 \leq y < 1$ (which is always true for $y$ in $[0,1)$).

Therefore, such $y$ exists, for example, $y = 0.97$.

But let's check if $S(x)$ can actually be $12345$ for $x = a + y = 196 + y$ with $y$ in $[0.96875, 1)$.

But recall that $[x] = a$ (since $0 \leq y < 1$), $[2x] = 2a + [2y]$, $[4x] = 4a + [4y]$, etc.
So,
$$S(x) = 63a + [y] + [2y] + [4y] + [8y] + [16y] + [32y]$$
For $a = 196$, $S(x) = 12348 + m$.
But $12345 = 12348 + m$ implies $m = -3$.
But $m$ cannot be negative, as $m = [y] + [2y] + [4y] + [8y] + [16y] + [32y] \geq 0$.

Wait, but above we had $12345 = 63 \times 196 + 57 = 12348 + 57 = 12405$.
But $63 \times 196 = 12348$, $12348 + 57 = 12405$.
But $12345 = 63 \times 195 + 60 = 12285 + 60 = 12345$.
So $12345 = 63 \times 195 + 60$.
So $m = 60$.
But $m$ can only be $0$ to $57$.
Therefore, $S(x)$ can only take values $63a + m$ for $0 \leq m \leq 57$.

Therefore, $12345$ cannot be written as $63a + m$ with $0 \leq m \leq 57$.

Therefore, the equation has no real solution.