And now, an attempt to prove that the given sequence will always give a result of 9...

1.Pick a number from 1 to 9

2.Multiply it for three

3.Plus 3

4.Multiply it for 3 again

Assume that the chosen number X is an integer between 1 and 9 inclusive. Going through the sequence step by step reveals an equation.

1. X

2. 3X

3. 3X+3 = 3(x+1)

4. 3*3(x+1) = 9(x+1)

As you can clearly see, regardless of what X is, the result at step 4 will always be divisible by 9. Since X is an integer between 1 and 9 inclusive, the result at step 4 must range between 18 and 90.

Now, there's a rule of divisibility for 9 where if the summation of the digits of a number is divisible by 9, then it too is also divisible by nine. Since the result at step 4 is a two digit number, the summation of it's digit cannot exceed 18 (9+9). What's more, no two single digit combination other than 9 and 9 sum to 18. Because 99 is outside of the range, but the result at step 4 is always divisible by nine, then the summation of the digits of the result from step 4 will always equal 9.