Le Monde puzzle [#752]

December 8, 2011
By

(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

After a loooong break, here is one Le Monde mathematical puzzle I had time to look at, prior to going to Dauphine for a Saturday morning class (in replacement of my R class this week)! The question is as follows:

A set of numbers {1,…,N} is such that multiples of 4 are tagged C and multiples of 5 and of 11 are tagged Q. Numbers that are not multiples of 4, 5, or 11, and numbers that are multiples of both 4 and 5 or of both 4 and 11 are not tagged. Find N such that the number of C tags is equal to the number of Q tags.

This is a plain enumeration problem.

N=0
noco=TRUE
nbC=nbQ=0

while (noco){
 N=N+1
 divF=FALSE
 if (trunc(N/4)*4==N){
    nbC=nbC+1
    divF=TRUE
    }
 if ((trunc(N/5)*5==N)||(trunc(N/11)*11==N)){
   if (divF){
     nbC=nbC-1
     }else{ nbQ=nbQ+1}
   }
 noco=(nbC!=nbQ)
 }

When I ran the code, I found many solutions

[1] 1 0 0
[1] 2 0 0
[1] 3 0 0
[1] 5 1 1
[1] 6 1 1
[1] 7 1 1
[1] 10  2  2
[1] 12  3  3
[1] 13  3  3
[1] 14  3  3
[1] 16  4  4
[1] 17  4  4
[1] 18  4  4
[1] 19  4  4
[1] 20  4  4
[1] 21  4  4
[1] 24  5  5
[1] 28  6  6
[1] 29  6  6
[1] 32  7  7
[1] 64 12 12

with no value further than 64 (testing all the way to 3,500,000). This seems in line with the fact that there are more multiples of 5 or 11 than of 4 when N is large enough. This can be seen by drawing the curves of the (approximate) number of multiples:

curve((trunc(x/4)-trunc(x/20)-trunc(x/44)),
  from=10,to=250,n=500)
curve((trunc(x/5)+trunc(x/11)-trunc(x/55)-
  trunc(x/20)-trunc( /44)),from=10,to=250,add=TRUE,n=500)


Filed under: R, Statistics Tagged: arithmetics, Le Monde, mathematical puzzle, number theory, Phoenix, R, Université Paris Dauphine, WSC 11

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