Carmichael number, more or less

This article is originally published at https://xianblog.wordpress.com

A quick-and-dirty R resolution of a riddle from The Riddler, namely to find a Carmichael number of the form abcabc:

library(numbers)
for(i in 1:9)
  for(j in 0:9)
    for(k in 0:9){
      x=i*100100+j*1010+k*101
      if(!isPrime(x)){
        p=primeFactors(x)
        if((prod(apply(outer(p,p,F="=="),1,sum)%%2))&
          (!max((x-1)%%(p-1))))break()}}

resulting into the number 101 101, since its prime factors are

> primeFactors(101101)
[1]   7  11  13 101

and 6, 10, 12, and 100 are divisors of 101100:

> primeFactors(101100) 
[1] 2 2 3 5 5 337

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This article is originally published at https://xianblog.wordpress.com
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