Question 228139
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


{{{(n(n+1)(n+2))*(1/(n+1))=2400(n+1)*(1/(n+1))}}}


{{{n^2+2n=2400}}}


Subtract 2400 from both sides of the equation


{{{n^2+2n-2400=2400-2400}}}


{{{n^2+2n-2400=0}}}


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


{{{n^2+2n-2400=(x+50)(x-48)=0}}}.  


This implies that


{{{x+50)=0}}} or {{{x=-50}}}


and


{{{x-48)=90}}} or {{{x=48}}}


We also could have use the quadratic formula given as


{{{n = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


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*49*50=800(48+49+50)}}}


{{{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.


{{{(-48)*(-49)*(-50)=800(-48+(-49)+(-50))}}}


{{{-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.


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Good luck in your studies!


Respectfully,
Dr J