Twitter Math Puzzle and Solution

July 7, 2011
By

(This article was first published on John Myles White » Statistics, and kindly contributed to R-bloggers)

Yesterday I posted a very simple math puzzle to Twitter that I found in Jonathan Baron’s book, Thinking and Deciding. The puzzle is the following:

Show that every number of the form ABC,ABC is divisible by 13.

The puzzle comes up in Baron’s book as an example of an “insight problem” in which one goes from not knowing the answer at all to knowing the complete answering in a sudden moment of insight.

Several people replied to my tweet with solutions: I especially like Will Townes’s solution. In particular, if you’re familiar with modular arithmetic, I like the logic of Will’s answer because it gives a simple generalization. First, represent ABC,ABC as ABC * 1000 + ABC * 1 rather than as ABC * 1001. Then notice that

  1. 1 = 1 mod 13
  2. 1000 = -1 mod 13

Thus ABC,ABC = ABC * -1 + ABC * 1 = 0 mod 13. This logic can be easily extended to show that (ABC,ABC,)*ABC,ABC = 0 mod 13 no matter how many times you repeat the ABC,ABC pattern.

To leave a comment for the author, please follow the link and comment on their blog: John Myles White » Statistics.

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Twitter Math Puzzle and Solution

July 7, 2011
By

(This article was first published on John Myles White » Statistics, and kindly contributed to R-bloggers)

Yesterday I posted a very simple math puzzle to Twitter that I found in Jonathan Baron’s book, Thinking and Deciding. The puzzle is the following:

Show that every number of the form ABC,ABC is divisible by 13.

The puzzle comes up in Baron’s book as an example of an “insight problem” in which one goes from not knowing the answer at all to knowing the complete answering in a sudden moment of insight.

Several people replied to my tweet with solutions: I especially like Will Townes’s solution. In particular, if you’re familiar with modular arithmetic, I like the logic of Will’s answer because it gives a simple generalization. First, represent ABC,ABC as ABC * 1000 + ABC * 1 rather than as ABC * 1001. Then notice that

  1. 1 = 1 mod 13
  2. 1000 = -1 mod 13

Thus ABC,ABC = ABC * -1 + ABC * 1 = 0 mod 13. This logic can be easily extended to show that (ABC,ABC,)*ABC,ABC = 0 mod 13 no matter how many times you repeat the ABC,ABC pattern.

To leave a comment for the author, please follow the link and comment on their blog: John Myles White » Statistics.

R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...



If you got this far, why not subscribe for updates from the site? Choose your flavor: e-mail, twitter, RSS, or facebook...

Tags: , ,

Comments are closed.

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