Le Monde puzzle [41]

The 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

η(160)+5=5η(2)+η(5)=15

η(1809)+3=3η(3)+η(67)=10

Since η(2) cannot be 1 and is an integer, we necessarily have η(2)=2 and this implies η(5)=5. With the same constraint on 1, the second sum also leads to the unique solution η(3)=2 and η(67)=4. The solution is thus η(2010)=13-3=13. (For ‘Og readers who saw several versions of this post, the subsidiary question is how many versions was there?!)

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