On a table, there are 100 piles of pebbles, containing 1, 2, ..., 100 pebbles respectively. In one step, we can reduce any number of selected piles, provided that we remove the same number of pebbles from each.

In the least number of steps, how can we remove all the pebbles from the table?
First, we prove that this can be done in seven steps; for example, in the following way.

1. Step: From the piles containing more than 63 pebbles, we remove 64. Thus, in these, at most 36 remain, and in the others, at most 63, so there is no pile with more than 63 pebbles.
2. Step: Where possible, we remove 32 pebbles. By a similar line of thought as before, then in each pile, there can be at most 31 pebbles.
3. Step: We take 16 pebbles from the appropriate piles. Then, at most 15 can remain everywhere.

From here, the method is already visible: in the 4th step, by taking 8, at most 7 can remain everywhere; in the 5th step, by taking 4, at most 3 can remain; then by taking 2, at most 1 can remain everywhere; and in the 7th step, by taking this one, we are done.

Now we see that this could not have been done in six steps. Suppose it could: in the six steps, taking $a_{1}, a_{2}, \ldots, a_{6}$ pebbles from certain piles, everything was emptied by the end.

Assign to each pile a code number of at most six digits in binary as follows. The $i$-th digit of the number $(i=1,2, \ldots, 6)$ should be 1 if we took pebbles from the pile in the $i$-th step (then only $a_{i}$ could have been taken); and 0 if not. Thus, we have assigned a unique number to each pile. At the same time, such a number can only belong to one pile, since the following decoding procedure is valid. The number $\overline{x_{6} x_{5} x_{4} x_{3} x_{2} x_{1_{2}}}$ means that in the first step, we took $x_{1} \cdot a_{1}$ pebbles (where $x_{1}$ can be 0 or 1), in the second step $x_{2} \cdot a_{2}$, so in the six steps, we took a total of $x_{1} a_{1}+x_{2} a_{2}+\ldots+x_{6} a_{6}$. The final result, however, is 0, so there were originally $x_{1} a_{1}+\ldots+x_{6} a_{6}$ pebbles in the pile; from the code, we can thus tell which pile it belongs to.

Since we started with 100 piles of different sizes, we would have needed 100 different codes. However, there are only 64 different six-digit numbers in binary (and, for example, all zeros cannot even be a code); this contradicts our previous observation, so it is not possible to remove all the pebbles in six steps.