Alok Goyal's Puzzles Page

Many correct responses to this puzzle, but the I was happiest to see was Danny from Washington DC, an 8th grader – well done Danny. Other people who sent the right answer included Suman Saraf, Vivek Rai and Pratik Poddar. Thank you all!

The answer is that both Jack and Will will have the same number of coins at the end of the game. The reason is that they pass on the same number of coins to each other in every turn. Here is how this works (copying the answer from Pratik as well as Danny):

(J,W) = (2015!, 2015!) –> (2015!/2, 3/2*2015!) –> (2015!, 2015!) –> (2015!3*/4, 5/4*2015!) –> (2015!, 2015!) and so on

In other words, say Jack and Will have X amount of gold.  Jack gives Will X/n amount of gold, so Will ends up with X +X/n  OR (n+1)X/n amount of gold. Now Will gives Jack the amount he has divided by (n+1)  back , which equals X/n. 

Note that if instead of 2015, we had an even number, the answer would not be same!

Hope you all enjoyed the puzzle!

Apologies for a long period of silence, I was traveling on work and then on vacation outside India and in the process just did not have the time to respond to messages and also add any posts. Normal service resumes though!

I only got three correct answers for this puzzle – two of them being 9 year olds and third one 11 year old, I presume it was too simple for most people. Congratulations to Mahi Saraf, Arushi Goyal and Aneyaa Dube!.

In the interest of time, I will keep the explanation short for this answer, though it is important to note that this method is often used for many similar magic grid puzzles. Here is an explanation, and I am also attaching the full solution from Mahi Saraf as well:

• If we add all the numbers, they sum to 36, where as if we add all the 4 lines (each of which add up to 12), they add to 48. That leads to the conclusion that the center 4 boxes add up to 12
• Center 4 boxes, therefore, can be 1,2,3,6 or 1,2,4,5
• Recognize that 8 needs to be in outer boxes – whichever line connects to it, needs to have the other two boxes add up to 4 (12 – 8). If the center boxes had 1,2,4,5, we cannot get any combination of two boxes that add up to 4. Therefore, the center boxes will have 1,2,3,6
• We also now know that 1 and 3 need to be adjacent to each other
• We also know that 1 and 2 cannot be adjacent to each other as that line will need a “9” to add up to 12 then, therefore, 1 and 2 need to be opposite to each other and 6 and 3 will be adjacent to each other
• You can now form the full picture

Puzzle 143 Solution

Mahi Saraf Solution Part 1

Mahi Saraf Solution Part 2

Hope you all enjoyed the puzzle!