Question

Find four consecutive integers such that the sum of the first three is 44 more than the fourth.

Answer 1

We will have the following:

We will have to sum 3 consecutive numbers, so we will name those numbers:

n

n+1

n+2

and the fourth number n+3

We can see that they are consecutive since the sepatation is of 1 unit each, we will also have that they must follow:

$n+(n+1)+(n+2)=(n+3)+44$

So, we solve for n:

$\Rightarrow3n+3=n+47\Rightarrow2n=44$$\Rightarrow n=22$

So, the three consecutive numbers are: 22, 23, & 24:

And we prove this by adding them and then subctacting 44 nd we should get 25:

$22+23+24=69-44=25$