the average ( arithmetic mean ) of 4 different positive integers is 16 . if the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer , what is the least possible value of the fourth integer ?	Choose the correct option.
A) 5  B) 4  C) 3  D) 2  E) 1

Rationale: let the second integer be x and the fourth be a . then [ 3 x + x + ( x + 2 ) + a ] / 4 = 16 = > 5 x + 2 + a = 64 = > 5 x + a = 62 = > a = 62 - 5 x from the above equation we can see that a is minimum when x is maximum , provided both are positive the maximum value that x can take in the above equation while still keeping a positive is x = 12 this gives us a = 62 - 60 = 2 therefore the minimum value that the fourth integer can have is 2 . option d .
Answer: D