The average (arithmetic mean) of four 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 answer.
A) 5  B) 4  C) 3  D) 2  E) 1
Let the second integer be x and the fourth be a.   Then [3x + x + (x+2) + a]/4 = 16   =&gt; 5x + 2 + a = 64   =&gt; 5x + a = 62   =&gt; a = 62 - 5x   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= 
<gadget id="calculator">62 - 60</gadget>
<output>2</output>
2   Therefore the minimum value that the fourth integer can have is 2. Option 2.
<result>D</result>
The average (arithmetic mean) of four 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?