There are no two consecutive integers which sum to make 48. The closest you can get is if you drop the integer part of the problem and state simply that they have to have a difference of 1 between them. In which case it would be 23.5 and 24.5.
48 1/3, 48 2/3, 48 3/3
48 = 6 x 8, so the smaller is 6.
Suppose the first number is 2x (rather than x, so that you force it to be even), where x is an integer. Then the next (consecutive) even number is 2x+2. So the equation is: 2x * (2x+2) = 48
15, 16 and 17
14,16.18
48 1/3, 48 2/3, 48 3/3
Because 6*8 = 48 and 48/8 = 6
48 = 6 x 8, so the smaller is 6.
-17, -16 and -15
The integers are 44, 46 and 48.
Suppose the first number is 2x (rather than x, so that you force it to be even), where x is an integer. Then the next (consecutive) even number is 2x+2. So the equation is: 2x * (2x+2) = 48
15, 16 and 17
14,16.18
No solution. 22 + 24 = 46 24 + 26 = 50
The numbers are -18, -16 and -14.
There is no set of four consecutive odd integers for 204. The only set is even: 48, 50, 52 and 54.
They are 6 times 8 = 48