# 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: