The average (arithmetic mean) of four different positive integers is 4. If the first of these integers in 3 times the second integer and the second integer is 6 less than the third integer, what is the least possible value of the fourth integer?
Choose the most appropriate option
A) 5
B) 3
C) 4
D) 2
E) 1
Let the second integer be x and the fourth be a.   Then [3x + x + (x+2) + a]/4 = 9   =&gt; 5x + 2 + a = 36   =&gt; 5x + a = 34   =&gt; a = 34 - 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=6   This gives us a= 34 - 30= 
<gadget id="calculator">34 - 30</gadget>
<output>4</output>
4   Therefore the minimum value that the fourth integer can have is 4.   Answer:4
<result>C</result>
The average (arithmetic mean) of four different positive integers is 4. If the first of these integers in 3 times the second integer and the second integer is 6 less than the third integer, what is the least possible value of the fourth integer?