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From: R-bloggers <[email protected]>
Date: Mon, Jun 17, 2019 at 6:13 PM
Subject: [R-bloggers] Le Monde puzzle [#1104]
To: <[email protected]>


[R-bloggers] Le Monde puzzle [#1104] <https://www.r-bloggers.com> [image:
Link to R-bloggers] <https://www.r-bloggers.com>
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Le Monde puzzle [#1104]
<http://feedproxy.google.com/~r/RBloggers/~3/Dhdp0uFh-WA/?utm_source=feedburner&utm_medium=email>

Posted: 17 Jun 2019 03:19 PM PDT

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(This article was first published on * R – Xi'an's Og
<https://xianblog.wordpress.com/2019/06/18/le-monde-puzzle-1104/>*, and
kindly contributed to R-bloggers) <https://www.r-bloggers.com/>

*A* palindromic Le Monde mathematical puzzle
<https://xianblog.wordpress.com/2011/09/03/le-monde-puzzle-website/>:


*In a monetary system where all palindromic amounts between 1 and 10⁸ have
a coin, find the numbers less than 10³ that cannot be paid with less than
three coins. Find if 20,191,104 can be paid with two coins. Similarly, find
if 11,042,019 can be paid with two or three coins. *

Which can be solved in a few lines of R code:

coin=sort(c(1:9,(1:9)*11,outer(1:9*101,(0:9)*10,"+")))
amounz=sort(unique(c(coin,as.vector(outer(coin,coin,"+")))))
amounz=amounz[amounz<1e3]

and produces 10 amounts that cannot be paid with one or two coins. It is
also easy to check that three coins are enough to cover all amounts below
10³. For the second question, starting with n¹=20,188,102,  a simple
downward search of palindromic pairs (n¹,n²) such that n¹+n²=20,188,102 led
to n¹=16,755,761 and n²=3,435,343. And starting with 11,033,011, the same
search does not produce any solution, while there are three coins such that
n¹+n²+n³=11,042,019, for instance n¹=11,022,011, n²=20,002, and n³=6.

To *leave a comment* for the author, please follow the link and comment on
their blog: * R – Xi'an's Og
<https://xianblog.wordpress.com/2019/06/18/le-monde-puzzle-1104/>*.
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