SOLUTION: Which three consecutive integers have a product that is 800 times their sum?

Question 228139: Which three consecutive integers have a product that is 800 times their sum?
Found 2 solutions by Alan3354, drj:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Which three consecutive integers have a product that is 800 times their sum?
x*(x+1)*(x+2) = 800*(3x+3)
x^3 + 3x^2 + 2x = 2400x + 2400
x^3 + 3x^2 - 2398x - 2400 = 0
x = -1
Divide by (x+1)
--> x^2 + 2x - 2400 = 0
(x + 50)*(x - 48) = 0
x = 48
x = -50

Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
Which three consecutive integers have a product that is 800 times their sum?

Step 1. Let n be the first integer.

Step 2. Let n+1 and n+2 be the next two consecutive integers.

Step 3. Let n+n+1+n+2=3n+3=3(n+1) be the sum of the three consecutive integers.

Step 4. Let n(n+1)(n+2) be the product of the three consecutive integers

Step 5. Then,n(n+1)(n+2)=800*3(n+1) since the product in Step 4 is 800 times their sum in Step 3.

Step 6. Solving n(n+1)(n+2)=800*3(n+1) yields the following steps:

Simplify by dividing n+1 to both sides of the equation

Subtract 2400 from both sides of the equation

Step 7. To solve, we can factor as follows:

.

This implies that

or

and

or

We also could have use the quadratic formula given as

where a = 1, b=2, and c=-2400.

Step 9. With n=48, then n+1=49 and n+2=50. Check the relationship in Step 5 n(n+1)(n+2)=800(n+n+1+n+2) with these values.

which is a true statement.

Step 10. With n=-50, then n+1=-49 and n+2=-48. Check the relationship in Step 5 n(n+1)(n+2)=800(n+n+1+n+2) with these values.

which is a true statement.

Step 8. ANSWER: There are two sets of consecutive numbers. The first set is 48, 49, 50 and the second set is -48, -49, -50.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

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