UNSW School Mathematics Competition Problems 2011

Junior Division - Problems and Solutions
 
 
Problem 1
A second-cousin prime $n$-tuple is defined as a set of $n$ prime numbers  $\{p, p+6, \ldots p+6(n-1)\}$ with common difference six. Each number in the set is a prime and consecutive members of the set differ by six.
For example 2011 is a member of a second-cousin prime 2-tuple.
 
Show that there is one and only one second-cousin prime 5-tuple and there are no second-cousin prime 6-tuples.
 
Solution 1
Clearly if $p$ is not equal to five and is a member of a second-cousin prime $n$-tuple  then the last digit of $p$ must be one of one, three, seven or nine. Suppose it ends
in one, then the next member of the second-cousin prime $n$-tuple ends with a seven, the next member a three, the next member a nine and then the next number that differs by six ends in a five and
is therefore non-prime. Thus there are no second-cousin prime $n$-tuples with $n>4$ if the first prime in the set is not equal to five. It remains to consider a second-cousin prime $n$-tuple starting with $p=5$.
By construction the largest second-cousin prime $n$-tuple is the second-cousin prime 5-tuple $(5,11,17,23,29)$ and there are no second-cousin prime 6-tuples.