The average (arithmetic mean) of four different positive integers is 15. 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 choice.
A) 8
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 =15   =&gt; 5x + 2 + a = 60   =&gt; 5x + a = 58   =&gt; a = 58 - 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=10   This gives us a= 58 - 50= 
<gadget id="calculator">58 - 50</gadget>
<output>8</output>
8   Therefore the minimum value that the fourth integer can have is 8. Option 8.
<result>A</result>
The average (arithmetic mean) of four different positive integers is 15. 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?