Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 1H / ii / Solution

The prime factorization is $561=3\cdot 11\cdot 17$, so it is composite and squarefree. Moreover $3-1=2$, $11-1=10$ and $17-1=16$ all divide $560=561-1$. The Korselt criterion therefore proves that 561 is a Carmichael number.