The product of three consecutive integers is 120. That product divided by the mean of the three integers is 24. What is the largest of the three consecutive integers?
Let's call the first integer x. So the three consecutive integers are x, x+1, and x+2. The product of those integers is x * (x+1) * (x+2). That's correct. And if we divide that product by the mean of the three integers, we get x * (x+1) * (x+2) / ((x + x+1 + x+2) / 3). The mean of the three integers is just (3x+3)/3 = x+1. So the equation we need to solve is x * (x+1) * (x+2) / (x+1) = 24. That gives us x * (x+2) = 24. If we FOIL the left side, we get x^2 + 2x = 24. That's correct. We can subtract 24 from both sides and then use the quadratic formula to solve for x. The quadratic formula is x = (-b +- sqrt( b^2-4*a*c ))/(2*a). So let's plug in a=1, b=2, and c=-24. Doing so gives us x = (-2 +- sqrt( 2^2-4*1*(-24) ))/(2*1). That simplifies to x = (-2 +- sqrt( 4 + 96 )) / 2. So x = (-2 +- sqrt( 100 )) / 2. The square root of 100 is 10. And -2 plus or minus 10 is 8 or -12. So x is either 8 or -12. The three consecutive integers are either 4, 5, and 6, or -6, -5, and -4. So the largest integer is either 6 or -4. But the product of the three integers has to be 120. Right. Let's try 4, 5, and 6. The product of these three integers is 120. So the largest integer is 6. Yes.

