find three consecutive integers whose product is 161 larger than the cube of the smallest integer

Question

 I need to know how to write this problem into a equation that I can solve.

Answer 1

Hi Allie,

 

Three consecutive integers can be written as n, n+1, and n+2. The product of these equals 161 more than the cube of the smallest integer, which would be n. In equation form, this reads:

 

n(n + 1)(n + 2) = n³ + 161

 

or

 

n³ + 3n² + 2n = n³ + 161

 

or

 

3n² + 2n - 161 = 0, after canceling out the n³ terms.

 

(3n + 23)(n - 7) = 0

 

3n = 23 which does not lead to an integer, so can be discarded.

 

So, n = 7. The three consecutive integers are 7, 8, 9.

 

Answer 2

Let the 3 consecutive integers be

 

n (smallest)

n+1

n+2

 

Then from the problem statement,

 

n(n+1)(n+2) = n³ + 161

 

n(n² + 2n + n + 2) = n³ + 161

 

n³ + 3n² + 2n = n³ + 161

 

Note that n³ cancels out on both sides, so we do not have to solve a cubic equation!

 

3n² + 2n = 161 ⇒ 3n² + 2n - 161 = 0

 

n = (1/6)[-2 ± √(2² + 4*3*161)] = (1/6)(-2 ± 44)

 

Note that only with +44 do we get an integer.

 

n = (1/6)(42) = 7

 

The three integers are 7, 8, and 9.

 

You go ahead and test the solution now.

Answer 3

Let the smallest integer be x

Your consecutive integers are x, x+1, x+2

 

x(x+1)(x+2) = x³ + 161

 

x³+3x²+2x = x³+161

Subtract x³ from both sides

3x²+2x = 161

subtract 161 from both sides

3x² + 2x - 161 = 0

 

You now have a quadratic equation which you can

use to solve for the smallest integer.