Write an expression for the next 3 even numbers that are greater than P

Every even number is 2 away from the last.

0, 2, 4, 6, 8, 10, 12, 14 ... etc.

If we had an even number p, then the next three even numbers would be

p+2, p+4, and p+6.

(If we had an odd number p, then the next three even numbers would be

p+1, p+3, and p+5. I'm not sure if p is even is implied in the question. Technically the answer would be p - p mod 2 + 2, where p is an interger ... that gets into more technical function stuff, though.)

We don't know whether ' P ' itself is odd or even.

- - If ' P ' is even, then the next three [larger] even numbers are

P+2, P+4, and P+6.

- - If ' P ' is odd, then the net three [larger] even numbers are

P+1, P+3, and P+5.