1. In the four-digit number 4753 , three two-digit numbers are formed by successive pairs of digits $(47,75,53)$. Exactly two of these two-digit numbers are prime. Find all four-digit numbers in which all four digits are prime, and all three two-digit numbers formed by successive digits are prime.
SolUTION
Let $P$ represent such a four-digit number.
As the digits of $P$ must each be prime, the only choices of digits we can make are $2,3,5$ or 7 . It will be helpful to list the two digit prime numbers which use only these digits: $23,37,53$, and 73 .

The first two digits of $P$ must form a prime number when taken as a two-digit number (from left to right), so we should consider each of the four options in turn.
If the first two digits are 2 and 3 respectively, this means that the third digit must be 7 as the second and third digits must form a two-digit prime number and our only choice is 37 . The final digit must be 3 , as the third and fourth digits must also form a two-digit prime number and our only choice is 73. This gives 2373 as a possible value for $P$.
If we consider the other three options for the first two digits in turn, similar reasoning yields 3737, 5373 and 7373 as possible values for $P$.

In conclusion, we have four possible values for $P$ which are: $2373,3737,5373$, and 7373 .