find three consecutive integers whose product is 385 larger than the cube of the smallest integer

Question

find three consecutive integers whose product is 385 larger than the cube of the smallest integer

Answer 1

What an interesting question! I like it. Of course the trick is to set up some equations that represent the problem. Let's take it piece by piece.

 

First let's think about the three consecutive integers. For them to be consecutive is just for them to be right next to each other (or in other words, each differs from the previous by only 1). 3, 4, and 5 would be an example of three such integers. Since we don't know what our integers will actually be, let's just call the smallest of them x. Our three consecutive integers are then x, x+1, and x+2.

 

Next we give ourselves the equation. If the product is 385 larger than the cube of the smallest, then the cube of the smallest plus 385 must be equal to the product. This is simply restating the same fact in different words, but gives us a better idea of how to write it as an equation. We can then simply multiply out our three integers and set them equal to the smallest cubed plus 385, like so:

 

(x)(x+1)(x+2) = x³+385

 

From here, solving the equation should be something you can do on your own. However, let's take it a little farther anyway. Let's multiply out the left side:

 

(x)(x+1)(x+2)  = x³+3x²+2x = x³+385

 

Subtract the x³ and the 385 from both sides and we get:

 

3x²+2x - 385 = 0

 

Now factoring that looks kinda hard, so I just used the quadratic formula to solve it instead. It tells me that x = -35/3 or x = 11. Since we're looking for consecutive integers, we can remove that first answer. Now we're almost done. All we need to do is find our other two integers, which is easy, since we said from the get go that x was the smallest, the other two must be 12 and 13.

 

Let's double check just to make sure.

 

(11)(12)(13) = 1716 = 1331+385 = 11³+385

 

So it looks good!