Question 221142
Show the impossible for three consecutive integers to have a sum that is 200 more than the smallest integer.  


Step 1.  Let n be the integer


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


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


Step 4.  Let 200+n be 200 more than the smallest integer.


Step 5.  Equate steps 3 and 4.


{{{3(n+1)=n+200}}}


{{{3n+3=n+200}}}



Subtract n+3 from both sides of the equation


{{{3n+3-n-3=n+200-n-3}}}



{{{2n=197}}}


Divide by 2 to both sides of the equation


{{{2n/2=197/2}}}


{{{n=197/2}}}  This statement does not provide a whole integer n since 197 is odd and not divisible by 2.  


Step 6.  Note that 197/2 is not a whole number so it's impossible for three consecutive integers to have a sum that is 200 more than the smallest integer.  


I hope the above steps and explanation were helpful. 


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


Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs.


Respectfully, 
Dr J