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

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Which three consecutive integers have a product that is 800 times their sum?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

Question 228139: Which three consecutive integers have a product that is 800 times their sum?
Found 2 solutions by Alan3354, drj:
Answer by Alan3354(21605) About Me  (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) About Me  (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

%28n%28n%2B1%29%28n%2B2%29%29%2A%281%2F%28n%2B1%29%29=2400%28n%2B1%29%2A%281%2F%28n%2B1%29%29

n%5E2%2B2n=2400

Subtract 2400 from both sides of the equation

n%5E2%2B2n-2400=2400-2400

n%5E2%2B2n-2400=0

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

n%5E2%2B2n-2400=%28x%2B50%29%28x-48%29=0.

This implies that

x%2B50%29=0 or x=-50

and

x-48%29=90 or x=48

We also could have use the quadratic formula given as

n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

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.

48%2A49%2A50=800%2848%2B49%2B50%29

117600=117600 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.

%28-48%29%2A%28-49%29%2A%28-50%29=800%28-48%2B%28-49%29%2B%28-50%29%29

-117600=-117600 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