Logic Riddles for Chess Players (4)

by Arne Kaehler
5/23/2020 – In our fourth logic riddle for chess players we have to deal with a pawn who is not very good with numbers. Similar to the previous logic riddles, the chess theme again carries the logic riddle and should not be taken too seriously. You are more than welcome to write the answer to the riddle and to detail your approach to find the solution in the comments.

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The dumb pawn

 

Once there was a white pawn who suddenly realized that he was the last piece remaining on the chess board. After a while, he realized that you need other pieces to play a game.

So, he just wanted to go home. But walking home would take a whole day. He checked his wallet to see if he had enough ducats left to buy a train ticket, but he didn't.

In this moment two knights came along, a ♘ and a ♞, and they saw the sad pawn sitting on the square of the white king.

The ♘ asked him why he would not leave for home. The pawn told the knights about his situation and asked them if they would carry him home for the ducats he had left.

The ♘and the ♞ looked at each other. Then the ♘ turned to the pawn and said: "Well, this is not enough money for us to take you home, but how about a business offer? I want you to run to the other side of the chess board and then back to us, as fast as you can. If you manage to do so, I will double up the money you have! And you can run up and down the board as often as you want to!" The pawn listened carefully while the ♘ added:  "But, after each run you have to give exactly 24 Ducats to the ♞!" The ♞ looked a bit surprised but the ♘ continued: "Do we have a deal, pawn?"

"Indeed, we have ♘! Thank you so much!" said the pawn, very happy about the generous offer.

The pawn ran as fast as he could to the other side of the board and then back to the ♘♞s. Maybe he could make a lot of ducats like this!

When the pawn came back to the knights, the ♘ doubled the amount of ducats the pawn had, and with a smile the pawn gave 24 ducats to the ♞.

Immediately after this exchange of ducats, the pawn started to run to the other side of the chess board and back to the ♘♞s again. The second time he was a bit more exhausted.

The ♘ again doubled the money the pawn had and the pawn again gave 24 ducats to the ♞. The pawn then stopped for a moment to catch his breath and to count his money. It still was not enough for a train ticket.

But after the pawn came back the third time, gasping desperately for air, the knights started to laugh.

As soon as the pawn received double the amount of the money he had and gave 24 ducats to the ♞, he also realized why they were laughing: he had exactly zero ducats left.

But then both knights apologized for letting him run all the time and gave him all his money back. Then they took him home on their backs and the story came to a happy end.

The question is:

How many ducats did the pawn have before he started to run the first time across the chess board?

Solution of the "Slowest Knight" riddle

This puzzle was a bit different but did not require too much mathematical knowledge.

Most of you got it correct and I have to thank  JoshuaVGreen  for always being very engaged in the puzzles and coming with new ideas.

The best solution would be to exchange the knights, then take the bishop on g3 and try to liquidate into an endgame with the isolated pawn, while the...sorry...

I meant a very good solution is to swap the horses of the brothers!

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Arne Kaehler, a creative thinker who is passionate about board games in general was born in Hamburg and learned how to play chess at a very young age. Through teaching chess to youth teams and creating chess content on YouTube, Arne was able to extend this passion onto others and has even made an online chess course for anyone who wants to learn how to play this game. Currently, Arne blogs for the English news page of ChessBase and focuses on creating promotional and entertaining articles.
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haider33 haider33 5/25/2020 10:30
Kramnik ripped Carlsen apart in 2007 as he adopted the Botvinick system.
Flávio Patricio Doro Flávio Patricio Doro 5/24/2020 07:13
Yet another way to settle the problem:
Say the pawn had 24-x ducats at start. It had to be less than 24.
Well, the first run cost him x ducats, as he ended it with 24-2x ducats (48-2x-24).
Similarly, the second run cost him 2x ducats, and the third run 4x. So he lost 7x at the end.
Now, (24-x)-7x=0 returns x=3. So the pawn had 21 ducats at start (24-3=21).
Admittedly @DrBob64's solution is more precise than mine.
toni mateescu toni mateescu 5/24/2020 09:18
For a general form it can be used a(n) = 2^n *a(0) - 24(2^n-1). In this case n = 3.
Johannes Fischer Johannes Fischer 5/24/2020 07:41
@Metaphysician
Thanks for writing. The error was corrected.
Metaphysician Metaphysician 5/24/2020 06:11
The pawn is on e1, which is the original square of the white king, not the black king.
As stated below, working backwards, you can get to the answer, 21, with just simple arithmetic; no algebra required.
lajosarpad lajosarpad 5/24/2020 05:49
2(2(2x - 24) - 24) - 24 = 0

2(2(2x - 24) - 24) = 24

2(2x - 24) - 24 = 12

2(2x - 24) = 36

2x - 24 = 18

x - 12 = 9

x = 21
JoshuaVGreen JoshuaVGreen 5/23/2020 09:11
@DrBob64, I like your splitting things as 24 + x to remove the effect of the Black Knight's take. Very nice!
varunkulkarni varunkulkarni 5/23/2020 08:03
Solving it in reverse order, he had 12 left when he started the third run! So after finishing the same, doubling that made 24 and giving 24 to the Black knight made it exactly zero!

Going back another step, to end up with 12 ducats after all calculations of after completing the second run, he must have given 24 from double of what he had at the start of the second run. So the double of what he had would be the same as removing 24 from the doubled number and getting 12 i.e. 12+24=36. So he had half of that, 18 when he started the second run!

Similarly, to remain with 18 after the first run, he must have had 18+24=42 after his original amount was doubled. So his original amount was half of 42, which is 21!
DrBob64 DrBob64 5/23/2020 07:33
There are several ways to do this, including JVG's solution. A normal approach would be working backwards from the final state to get the starting amount. I like the following approach best.
Let the pawn start with 24 + x ducats. After one run, he has 24 + 2x ducats. After n runs, he has 24 + 2^n*x
where ^ represents raising to the power. Knowing that 24 + 8x = 0, gives us x=-3.
JoshuaVGreen JoshuaVGreen 5/23/2020 04:53
Suppose the pawn started with D ducats.

The pawn runs to the other side of the chessboard and returns. The White Knight doubles his money to 2D.
The Black Knight takes his cut, leaving the pawn with 2D - 24.
The pawn runs to the other side of the chessboard and returns. The White Knight doubles his money to 2(2D - 24) = 4D - 48.
The Black Knight takes his cut, leaving the pawn with (4D - 48) - 24 = 4D - 72.
The pawn runs to the other side of the chessboard and returns. The White Knight doubles his money to 2(2D - 24) = 2(4D - 72) = 8D - 144.
The Black Knight takes his cut, leaving the pawn with (8D - 144) - 24 = 8D - 168.

Alas, the pawn is now broke, so 8D - 168 = 0, hence D = 21.
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