Pick a number from 1-9.
Multiply by 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.