Question 1203985: The interiors of all unit squares of a 10 by 10 board are to be painted. Any two squares that have a side in common will be painted with a different colour of paint. What is the least number of colours that will be needed?
I am not satisfied with the existing solution on the Algebra.com website. It is far too simple and is not "mathematically" conveyed in any manner. Of course, this is a marvelous free service, so my feedback might be taken with distaste, and I apologise for that, but if someone can give a more mathematical solution, I would really appreciate it.
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website! .
There is a painting using two colors: it is the chessboard painting.
From the other side, one color does not produce any solution.
Hence, two colors is the least number of different colors that is needed.
It is a standard, traditional, complete and mathematically absolutely correct reasoning and a solution.
See similar solutions at this forum under these links
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1096043.html
https://www.algebra.com/algebra/homework/Length-and-distance/Length-and-distance.faq.question.1095728.html
This problem is known for hundreds of years. With a similar solution, it is written and placed
in hundreds of textbooks in Math in different languages, that are collections of entertainment problems.
So far, I never heard that somebody would be unsatisfied or expressed doubts in validity of this solution.
In opposite, the normal human reaction is admiration that so simple and so elegant solution is possible.
You are the first person who expresses your unsatisfaction (due to unknown reason). My congratulations ( ! )
Answer by MathLover1(20850)  (Show Source):
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