# Math Help - f(a)=100, f(124a)=700 -> f(3a)=?

1. ## f(a)=100, f(124a)=700 -> f(3a)=?

Let $f(n)$ denote the sum of the digits of the positive integer $n$ in decimal notation. What may $f(3a)$ be if $f(a)=100$ and $f(124a)=700$?

2. Hello, james_bond!

Is this a trick question?

Let $f(n)$ denote the sum of the digits of the positive integer $n$ in decimal notation.
What may $f(3a)$ be if $f(a)=100$ and $f(124a)=700$?

Anyone familiar with "Casting out 9's" know this fact . . .
. . If $f(n)$ is the digital-sum function: . $f(a\cdot b) \:=\:f(a)\cdot f(b)$

(Hmm, that's not quite accurate, but it'll do.)

If $f(a) = 100$, then: . $f(3a) \;=\;f(3)\!\cdot\!f(a) \;=\;3\cdot100 \;=\;300$ . . . . too easy

3. It wasn't supposed to be a trick question and I think that you're wrong since for example $f(9)f(9)=81\ne f(9*9)=f(81)=9$.

4. I know. a is the 100-digit number consisting of all 1’s, so 124a is the 102-digit number whose first two digits are 13 and last two 64, with 98 7’s in between. Hence f(3a) = 300 after all.

5. Originally Posted by JaneBennet
I know. a is the 100-digit number consisting of all 1’s, so 124a is the 102-digit number whose first two digits are 13 and last two 64, with 98 7’s in between. Hence f(3a) = 300 after all.
Very neat! But can you be sure that this is the unique solution???

6. Well, I’ve answered the question (what may f(3a) be) so I’m happy with that. There may be other answers, I don’t know, but I don’t think they are as neat as this simple one.