ok, here's a solution to 1):

using the fact that a-1 < [a] \leq a, the equation 1) gives us: 63x - 6 < 12345 \leq 63x. therefore [x] \in \{195,196 \}. now let x=[x] + \theta, where 0 < \theta < 1. then 1) will give us:

[2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=12345 - 63[x]. so we must have 12345 \geq 63 [x], which doesn't hold for [x]=196. hence we must have [x]=195 and therefore:

[2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=60. using the inequality [a] \leq a, we'll get: 60 \leq 62 \theta. thus: \frac{30}{31} \leq \theta < 1. therefore \frac{60}{31} \leq 2\theta < 2, and hence [2 \theta]=1. also \frac{120}{31} \leq 4\theta < 4 and so

[4 \theta]=3. similarly we see that [8 \theta]=7 and [16 \theta]=15. thus: 26 + [32 \theta]= [2 \theta] + [4\theta] + [8\theta] + [16\theta] + [32 \theta]=60, which gives us: [32 \theta]=34, which is impossible because 0 < \theta < 1

and hence [32 \theta] is at most 31. so the equation has no solution!