The sum of three consecutive number is 120 what is the smallest of the three numbers?

Let the numbers be 'n' , 'n+1' & 'n+2'

Hence adding

n + n+1 + n + 2 = 120

3n + 3 = 120

3n = 117

n = 39

Hence

n + 1 = 40

n + 2 = 41

So the three consecutive numbers are ;- 39,40,& 41.