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

**james_bond** · December 15th 2007, 02:32 AM

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 $f(124a)=700$ ?

**Soroban** · December 15th 2007, 03:35 AM

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

**james_bond** · December 15th 2007, 04:20 AM

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

**JaneBennet** · December 21st 2007, 06:07 PM

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.

**Opalg** · December 22nd 2007, 04:42 AM

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???

**JaneBennet** · December 22nd 2007, 05:30 AM

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.

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