The average (arithmetic mean) of four different positive integers is 12. 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
Let the second integer be x and the fourth be a.   Then [3x + x + (x+2) + a]/4 = 12   =&gt; 5x + 2 + a = 48   =&gt; 5x + a = 48   =&gt; a = 48 - 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=9   This gives us a= 48 - 45= 
<gadget id="calculator">48 - 45</gadget>
<output>3</output>
3   Therefore the minimum value that the fourth integer can have is 3. Option 3.
<result>C</result>
The average (arithmetic mean) of four different positive integers is 12. 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?