How do you find a set of four consecutive integers whose sum is equal to the sum of the next three consecutive integers immediately following them?
1 Answer
Explanation:
First let's identify the integers. The first one would be
#n+(n+1)+(n+2)+(n+3)=(n+4)+(n+5)+(n+6)#
Combine like terms.
#4n+6 = 3n + 15#
Now solve for
#4n-3n+6=cancel(3n-3n)+15#
#n+6=15#
Subtract
#ncancel(+6-6)=15-6#
#n=9#
We found that
#(9+1)=10#
#(9+2) = 11#
#(9+3)=12#
The other three integers are
Now we can check that these are the correct integers by finding the other three integers from the equation (the three integers that are the sum of the four known integers).
#(9+4)=13#
#(9+5)=14#
#(9+6)=15#
Now that we know the other three integers, we can plug all known values into the original equation and see if the integers solved for are correct.
#n+(n+1)+(n+2)+(n+3)=(n+4)+(n+5)+(n+6)#
#9+10+11+12=13+14+15#
#42 = 42#
The sums match, so the integers are: