JOMO 7, Short 4

Starting from 2 which is a prime number, all he does is to add 2, 0, 1 &3 repeatedly. Now when he adds 2, the number is divisible by 2, on adding 0 the number is still composite, when he adds 1, this makes the sum of digits divisible by 3, again on adding 3 the sum is again a multiple of 3. Hence there will be no prime number on repeating this cycle further. Hence we're left with 2 as the only prime number.

• Starting from 2, which is a prime number, we can make some assumptions:
• Every group of four digits "2013" can be divided for "3", because the sum of all digits will end in 3K , with K being an integer; so, if the number ends in "3", it'll never be a prime .
• If the number ends in "2" or "0", it'll be even, and obviously not a prime .
• If the number ends in "1" , it'll be followed by a "2" and lots of "2013", causing its sum to also be 3K; so, it'll will never be a prime .
• According to all this, the only prime number in the whole sequence will be number "2" .

If he adds $0$ to the end of the number the number is not prime as it is divisible by $2$ and $5$ .If he adds $2$ to the end of the number the number is not prime as it is divisible by $2$ .The cycle of $2013$ will always have been repeated a finite number of times before the number ends in $20$ so if he adds $1$ then $201$ is divisible by $3$ and the strings of $2013$ are always divisible by $3$ so the number is not prime.If he adds $3$ then a finite number of $2013$ s will be present in the number so it will be divisible by $3$ .So the only prime in in this sequence is $\boxed{2}$ so the answer is $\boxed{1}$ .

2 is the only prime number in this cycle, remaining all number are divisible by 2 or 3.