## Le Monde puzzle 

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)
 2 2 2 2 2 5
> prime.factor(1809)
  3  3  3 67
> prime.factor(2010)
  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?!)

This site uses Akismet to reduce spam. Learn how your comment data is processed.