Question 226398
What three consecutive numbers have a sum that is 1/5 of their product?


Step 1.  Let {{{n}}} be one number.


Step 2.  Let {{{n+1}}} and {{{n+2}}} be the next two consecutive numbers.


Step 3.  Let {{{n+n+1+n+2=3n+3=3(n+1)}}} be the sum


Step 4.  Let {{{n*(n+1)*(n+2)/5}}}  be {{{1/5}}} of their product


Step 5.  Then, {{{3(n+1)=n*(n+1)*(n+2)/5}}} since the three consecutive numbers have a sum that is 1/5 of their product.


Step 6.  Solving equation in Step 5, yields the following steps.


Divide {{{n+1}}} to simplify equation


{{{(n+1)*3/(n+1)=(1/(n+1))*(n*(n+1)*(n+2)/5)}}}


{{{3=n*(n+2)/5}}}


Multiply by 5 to get rid of denominator


{{{3*5=n*(n+2)}}}


{{{15=n^2+2n}}}


Subtract 15 from both sides of the equation


{{{15-15=n^2+2n-15}}}


{{{0=n^2+2n-15}}}


Step 7.  Factoring the above quadratic equation yields the following


{{{0=n^2+2n-15=(n+5)(n-3)}}}


Step 8.  So {{{n+5=0}}} and {{{n-3=0}}} or {{{n=-5}}} and {{{n=3}}}


Step 9.  With {{{n=-5}}}, {{{n+1=-4}}} and {{{n+2=-3}}}.  Check these numbers with the equation in Step 5 {{{3(n+1)=n*(n+1)*(n+2)/5}}}.  And


{{{-5-4-3=(-5)*(-4)*(-3)/5}}}


{{{-12=-12}}}  which is a true statement


Step 10.  With {{{n=3}}}, {{{n+1=4}}} and {{{n+2=5}}}.  Check these numbers with the equation in Step 5 {{{3(n+1)=n*(n+1)*(n+2)/5}}}.  And


{{{3+4+5=3*4*5/5}}}


{{{12=12}}} which is a true statement.


Step 11.  ANSWER:  There are two sets of consecutive numbers, they are 3, 4, 5 and -5, -4, -3.


I hope the above steps were helpful.


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Good luck in your studies!


Respectfully,
Dr J