Which 3 consecutive numbers when added together equal 9000 ?

2999, 3000, 3001.

Can we have a hard one now?

suppose the smallest is n

then n+n+1+n+2 =9000

ie 3n+3=9000

ie 3(n+1)=9000

so n+1=3000

so n=2999

so the numbers are 2999, 3000 and 3001

*It’s all about one’s concept of ‘consecutiveness’ so one can say 2998, 3000, 3002 as well because if 0.1 passes for consecutiveness then why not 2?

Anyway 1 should be the least controversial

Consecutiveness of squares too can be problematic as every number is the square of some number

I’m afraid, for the given number you can’t find consecutive natural squares.

Anyway if it were 8750 instead of 9000, then 2809, 2916 and 3025 will do!

*as referred to the following answer

2999+3000+3001=9000

2999, 3000, 3001 = 9000

3001 + 3000 + 2999

Hint: Divide by 3 and that will give you the average. Since they are three consecutive numbers they will be the average, one less and one more.

2999, 3000, 3001

You didn't specify that they had to be integers, as the previous responses all presumed.

2,999.9, 3000.00, and 3,000.1 are also consecutive, and solve the equation.

I would give "best answer" to someone who solves it with three consecutive Squares, though.

2,999 3000 3001

2999, 3000, 3001.