When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 intege | MathCelebrity Forum

When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 intege

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When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum?

Let the 3 consecutive positive integers be:
  1. x
  2. x + 1
  3. x + 2
The product is:
x(x + 1)(x + 2)

The sum is:
x + x + 1 + x + 2 = 3x + 3

We're told the product is equivalent to:
x(x + 1)(x + 2) = 16(3x + 3)
x(x + 1)(x + 2) = 16 * 3(x + 1)

Divide each side by (x + 1)
x(x + 2) = 48
x^2 + 2x = 48
x^2 + 2x - 48 = 0

Now subtract the sum from the product:
x^2 + 2x - 48 - (3x + 3)
x^2 - x - 51
 
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