SOLUTION: What is the formula for the following: The three least consecutive multiples of 8, the sum of which is greater than 244.

8n = first multiple of 8
8(n+1) = next multiple of 8
8(n+2) = third multiple of 8

Sum > 244

 8n + 8(n+1) + 8(n+2) > 244

8n + 8n + 8 + 8n + 16 > 244
             24n + 24 > 244
                  24n > 220
                    n > 
                    n > 
                    n > 

So the smallest possible value n could be is the
smallest integer greater than , which is 10.

So 

The first multiple of 8 = 8n = 8(10) = 80
The next multiple of 8 = 8(n+1) = 8(11) = 88
The third multiple of 8 = 8(n+2 = 8(12) = 96  

Checking: 80+88+96=254, so the sum is greater than 244.
That checks.

Checking to see if they are the least possible such 
consecutive multiples of 8.  To check this we try
the next lower consecutive multiples of 8.  They are
72, 80, 88, but their sum is only 240, so they are too
small, so 80, 88, and 96 are the smallest three consecutive
multiples of 8 whose sum exceeds 244.

Edwin