In a box, there are cards numbered with natural numbers from 1 to 2006. On the card numbered 2006 lies the card numbered 2005, and so on down to 1. In one move, it is allowed to take one top card (from any box) and place it either at the bottom of an empty box, or on a card with a number one greater. How many empty boxes are needed to transfer all the cards to another box?
The answer follows from the general fact: let the number of cards be $n$, where $2^{\mathrm{k}-1} \leq n < 2^{\mathrm{k}}$. If we assume that all cards can be placed in $k$ boxes, then card $a$ still needs to be moved so that all cards end up in one box. For this, at some point, it will be necessary to occupy the "top" cards of $k$ other boxes. In these boxes, "bottom" cards cannot be present (since directly under a "top" card, there can only be card $a$, which is still in box $i$). This means that all "bottom" cards are already in one box -- a contradiction. Therefore, it will require at least $k+1$ boxes.

## Answer

11 boxes.

## Problem