Rainbow Building

Let $R$ be the number of all discorded cubes, $r_{1}$ be the number of discarded cubes in the first process, $r_{2}$ be the number of discarded columns in the second process, and $r_{3}$ be the number of discarded bases in the third process.

Then $R = r_{1} + 3 r_{2} + 15 r_{3}$ .

Now for the original amount of cubes $N$ , we can write the modular operations as the following:

$N\equiv r_{1} \pmod 3$

$\dfrac{N - r_{1}}{3} \equiv r_{2} \pmod 5 \Rightarrow N - r_{1} \equiv 3r_{2} \pmod 5 \Rightarrow N \equiv r_{1} + 3r_{2} \pmod 5$

$\dfrac{\dfrac{N - r_{1}}{3} - r_{2}}{5} = r_{3} \pmod 7 \Rightarrow N - r_{1} - 3r_{2} \equiv 15r_{3} \pmod 7 \Rightarrow N \equiv r_{1} + 3r_{2} + 15r_{3} \pmod 7$

Now when we apply the modular operations upon $R$ , it will yield:

$R \equiv r_{1} \pmod 3$

$R \equiv r_{1} + 3r_{2} \pmod 5$

$R \equiv r_{1} + 3r_{2} + 15r_{3} \pmod 7$

Clearly, both $N$ & $R$ have the same remainders in the three moduli, and $lcm(3, 5, 7) = 105$ . Hence, with the relatively prime moduli, $N \equiv R \pmod {105}$ , according to Chinese Remainder Theorem .

From the question, $R = 52$ , and $N \equiv 0 \pmod {11}$ . As a result, we can substitute $N = 52 + 105m$ for some positive integer $m$ .

Thus, $N = 52 + 105m \equiv 0 \pmod {11}$ .

$-3 + 6m \equiv 0 \pmod {11}$

$6m \equiv 3 \pmod {11}$

$2m \equiv 1 \pmod {11}$

The smallest possible $m = 6$ . As a result, the least possible $N = 52 + 6\times 105 = \boxed{682}$ .

Let C be the number of cuboids that are contained in the final structure.

All cuboids are composed of 7 bases of 5 columns of columns containing 3 cubes.

Hence $N = 7\times 5\times 3\times C+52\\$

Assuming the final structure contains at least 1 complete cuboid, $N\ge 105\\$ and hence, $C\ge 1\\$

As N is a multiple of 11, $N=105C+52\equiv 0\quad (Mod\quad 11)\\$

$105C+52\equiv 0\quad(Mod\quad 11)\\ \Rightarrow 6C+8\equiv 0\quad(Mod\quad 11)\\ \Rightarrow 6C\equiv 3\quad(Mod\quad 11)\\ \Rightarrow C\equiv 6\quad (Mod\quad 11)$

Substituting back into the original equation, we have $N=105\times 6+52=\boxed { 682 } \\$