find four consecutive numbers such that the sum of the first three numbers is twelve more than the fourth number ?  Pick: A) 7,8,9,10 B) 3,4,5,6 C) 9,10,11,12 D) 6,7,8,9 E) None
Let   1st Integer = x   2nd Integer = x+1   3rd Integer = x+2   4th integer = x+3   =======================================   Given   x+x+1+x+2=(x+3)+12   3x+3=x+3+12   3x+3=x+15   3x=x+15-3   3x=x+12   3x-x=12   2x=12   x=12/2= 
<gadget id="calculator">12/2</gadget>
<output>6</output>
6========================================   1st Integer = x = 6   2nd Integer = x+1 = 6+1= 
<gadget id="calculator">6+1</gadget>
<output>7</output>
7   3rd Integer= x+2 =6+2= 
<gadget id="calculator">6+2</gadget>
<output>8</output>
8   4th integer= x+3=6+3= 
<gadget id="calculator">6+3</gadget>
<output>9</output>
9   Answer 6,7,8,9
<result>D</result>
find four consecutive numbers such that the sum of the first three numbers is twelve more than the fourth number ?