You begin with

At each step, you may replace aₙ with either:

1. 2aₙ + 1

2. ⌊aₙ / 2⌋ (the floor of half of aₙ)

The question:

Can you reach 2025?

This looks innocent. Double-and-add-one moves tend to grow numbers quickly; dividing by two seems to give you a way to “adjust.” So at first glance, maybe you can hop around and eventually land on 2025.

But reachability problems are rarely about brute force. They’re about structure.

Understanding the Moves

Let’s inspect both operations:

Move 1: 2a + 1

This is the operation that takes a number, shifts its binary digits left, and appends a 1 at the end.

Example:
13 → 27
(1101 → 11011)

This move always creates an odd number.

Move 2: floor(a / 2)

Binary right-shift by one (drop last bit).

Example:
13 → 6
(1101 → 110)

So the two moves let you append 1’s to the right, or chop off rightmost bits.

Now what is the binary of 2025?

Now try to think in binary and answer the question…

As usual, knight meets a dragon, and dragon says:

I will think of three “digits” a,b,c and you tell me three “numbers” x,y,z. Then I will compute and return the number ax+by+cz. Using this number, you have guess my digits a,b,c. If you fail, I will eat you.

Now the question is: in the place of the knight, what numbers would you say?

For example, if you say x = 12, y = 53, z = 7. Then the dragon would return you:

12a + 53b + 7c, now you have to figure out a,b,c. Sometimes there can be several possible options, but you have a single try :)

This is a tricky questions, but when you find out the answer, it turns out to be a silly question.

So, the best option would be x = 100, y = 10, z = 1. Because since the dragon has digits, he would return:

For example, if the dragon had a = 1, b = 5 , c = 4. The the result would be:

Then you can easily call the digits of the returned number, 1, 5, 4. That simple:)

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to group the given data according to some features and assign colors to groups.

If you are a data scientist or have studies some basic data science course, then this should sound familiar. For better understanding, let’s see some fundamental demo problems.

1.Bugs

Problem:

In each cell of the 5x5 board there is a ladybug. At some point, all the ladybugs crawl onto adjacent (horizontal or vertical) cells. Will this necessarily leave an empty cell?

Beetles are big for illustration purposes only. It is assumed that one cell could fit several beetles.

Solution:

First, if you pay close attention to where a ladybug can crawl, that should give some idea.

If the picture above didn’t ring a bell, then maybe imagine it like this:

So, if the board was painted in black and white, like a chessboard, then a ladybug sitting on a white cell can crawl to a black cell only, and a ladybug sitting on a black cell to a white cell only. Okay, this seems to be some an extra information about our problem. And we are obtaining this extra piece of information by visualizing the 5x5 board as “chessboard”. Then let’s re-state our problems with this extra detail:

In each cell of the 5x5 chessboard, which is painted in black and white colors, there is a ladybug. At some point, all the ladybugs crawl onto adjacent (horizontal or vertical) cells. Since cells adjacent to white cells are black and cells adjacent to black cells are white, ladybugs that are sitting on black cells will crawl to white cells and from white to black. Will this necessarily leave an empty cell?

So, at this point, just counting ladybugs by cell colors, should make the answer obvious.

12 ladybugs are sitting on white cells, 13 on black cells. What will we observe when the crawling happens? 13 ladybugs that were sitting on black cells need to move to white cells, number of which is only 12. So there must be several ladybugs in one the white cells. Similarly, 12 ladybugs that were sitting on white cells need to move to black cells, number of which is 13. These 12 ladybugs can’t fill 13 black cells, there will always be 1 empty black cell.

Answer:

There will always be 1 empty black cell.

2.Tetrominoes

Problem:

Is it possible to cover 10 by 10 chessboard by 25 straight tetrominoes?

Solution:

Since 10 is not divisible by 4 we encounter difficulties. Here you can get your pen and paper or in your imagination, try to do the covering. 10x10 has 100 cells and 25 straight tetrominoes have 100 cells. You need to perfectly cover all 100 cells of the 10x10 board, without overlapping, without breaking tetrominoes or using more than 25, etc…

I will go ahead and give you the answer, it is impossible. But how do you prove that it is impossible?

Here a simple coloring trick comes to help: let’s color the 10x10 board with 4 colors in the following way:

This coloring has a very interesting property: No matter how you place the tetromino on the board, it will always cover one of each of colors.

If we assume that it is possible to cover the board with 25 tetrominoes, then we have to cover with 25 tetrominoes, 25 of each colors, 25 red, 25 white, 25 blue, 25 green.

However, if we look at the board above, there are 26 red cells, which means we need at least 26 tetrominoes to cover the 10x10 board. So, it is not possible to do it with 25 tetrominoes.

The ideas above are like building blocks of larger and complex challenges. It takes some effort to be able to use them in your large scale applied tasks.