Try this:

Pick a number from 1-9.

Multiply by 3.

Add 3.

Multiply by 3 again.

Now add the two digits together.

The result is always 9!

Can anyone explain why this is so?

you always get a number dividable by 9

and all numbers dividable by 9 have that in them

(the +3 thing is just to make sure you get 2 digits

it would be the same as pick a number from 2 to 10 do times 9)

all numbers dividable by 9 can do it

you just need to add the digits over and over again till you get 1 digit

9279 ... 9+2+7+9 ... 27 ... 2+7 ... 9

Let N be your chosen digit, between 1 and 9. Then 3(3N+3) = 9N+9 is the result after your manipulations.

Now, consider an arbitrary two digit number, "ab." The value is 10a+b, and the sum of the digits is a+b. Can we write the result above in this form?

9N + 9 = (10N - N) + 9 = 10N + (9-N)

Therefore, a = N, b = 9-N.

The sum a+b = N + (9-N) = 9.

QED.