MHB What Is the Sum of the Digits of 3n Given These Conditions?

[anemone]

Here is this week's POTW:

-----

A number $n$ has sum of digits 100, while $44n$ has sum of digits 800. Find the sum of the digits of $3n$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulatons to kaliprasad for his correct solution!(Cool)

Here is a suggested solution by other:

Spoiler

Suppose $n$ has digits $a_1a_2\cdots a_k$ and digit sum $s=\sum a_i$. If we temporarily allow digits larger than 9, then $4n=(4a_1)(4a_2)\cdots(4a_k)$. Each carry of one reduces the digit sum by 9, so after making all necessary carries, the digit sum for $4n$ is at most $4s$. It can only be $4s$ iff there are no carries. Similarly, $11n$ has digit sum at most $2s$, with equality iff there are no carries.

Since $44n$ has digit sum $8s$, there cannot be any carries. In particular, there are no carriers in forming $4n$, so each digit of $n$ is at most 2. Hence, there are no carries in forming $3n$ and the digit sum of $3n$ is 300.