We have three integers, and we know they are all sequential integers, so we can refer to them as: x, x + 1, x + 2 We are also told: 3(x + 2) = x + (x + 1) + 47 Now we can use that relationship to solve for x: 3x + 2 = 2x + 48 x = 46 So our numbers are 46, 47 and 48
12 and 4 or 6 and 8
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.
-1
8 and 6 8 x 6=48 8 + 6=14
-46
We have three integers, and we know they are all sequential integers, so we can refer to them as: x, x + 1, x + 2 We are also told: 3(x + 2) = x + (x + 1) + 47 Now we can use that relationship to solve for x: 3x + 2 = 2x + 48 x = 46 So our numbers are 46, 47 and 48
12 and 4 or 6 and 8
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.
Because 6*8 = 48 and 48/8 = 6
Divide the sum of the three consecutive odd integers by 3: -147 /3 = -49. The smallest of these integers will be two less than -49 and the largest will be two more than -49, so the three consecutive odd integers will be -47, -49, and -51.
-1
8 and 6 8 x 6=48 8 + 6=14
48 = 6 x 8, so the smaller is 6.
997
The largest is 73.
26 is the largest