The average (arithmetic mean) of four different positive integers is 13. 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?
Select:
A) 4  B) 3  C) 2  D) 1  E) 0
Let the second integer be x and the fourth be a.   Then [3x + x + (x+2) + a]/4 = 13   =&gt; 5x + 2 + a = 52   =&gt; 5x + a = 50   =&gt; a = 50 - 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=50 - 50= 
<gadget id="calculator">50 - 50</gadget>
<output>0</output>
0   Therefore the minimum value that the fourth integer can have is 0. Option 0
<result>E</result>
The average (arithmetic mean) of four different positive integers is 13. 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?