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**T**he current puzzle in * Le Monde* this week is again about prime numbers:

The control key on a credit card is an integer η(a) associated with the card number a such that, if the card number is c=ab, its key η(c) satisfies η(c)=η(a)+η(b)-1. There is only one number with a key equal to 1 and the keys of 160 and 1809 are 10 and 7, respectively. What is the key of 2010?

The key of 1 is necessarily 1 since

*η(a1)=η(a)+η(1)-1=η(a).*

So this eliminates 1. Now, the prime number decompositions of 160, 1809, and 2010 are given by

> prime.factor(160)

[1] 2 2 2 2 2 5

> prime.factor(1809)

[1] 3 3 3 67

> prime.factor(2010)

[1] 2 3 5 67

using a function of the (still bugged!) schoolmath package. We thus have the decompositions

*5η(2)+**η(5)=11*

*3**η(3)+**η(67)=8*

Since *η(2)* cannot be 1 and is an integer, we necessarily have *η(2)=2* but then this implies *η(5)=1* (!), unless 0 is also a valid key (?), which would imply that *η(5)=11*. With the same constraint on 1, the second sum also leads to the unique solution *η(3)=2* and *η(67)=2.* I thus wonder if there is a bug in the key of 160…

Filed under: R, University life Tagged: Le Monde, mathematical puzzle, schoolmath

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