# Recreational mathematics - Digit sum

Find the sum of all 3-digit positive numbers N that satisfy the condition that the digit sum of N is 3 times the digit sum of N+3

Can you help me with this question?

-
Do you allow zero as leading digits – Amr Nov 27 '12 at 12:03
Your formatting is extremely weird: why use bold italic throughout all the question? Plus, it sounds like you're giving us an order. People here usually downvote for that ) A good way to avoid this is to follow the rules from the FAQ when asking questions. – Dan Shved Nov 27 '12 at 12:17
Numbers starting with 0 is not considered a 3digit number – chndn Nov 29 '12 at 10:30
SORRY I am new and i will edit this – chndn Nov 29 '12 at 10:31

Represent $N=100a+10b+c$. From the condition it is clear that $7\leq c\leq 9$ (otherwise the sum of digits in $N+3$ is $a+b+c+3$, which is greater than $a+b+c$).
Now we have two cases: $b<9$ and $b=9$.
(1) $b<9$ then the sum of digits in $N+3$ is $a+(b+1)+j$ where $j=0,1,2$. We have the equation: $$3(a+(b+1)+j)=a+b+c\Leftrightarrow 2a+2b+3j+3=c$$ Since $c=7,8,9$, we have three possible options: $$\begin{array}{l} c=7,j=0\hspace{5pt}\Rightarrow\hspace{5pt}2a+2b=4\hspace{5pt}\Rightarrow\hspace{5pt}a+b=2\\ c=8,j=1\hspace{5pt}\Rightarrow\hspace{5pt}2a+2b=2\hspace{5pt}\Rightarrow\hspace{5pt}a+b=1\\ c=9,j=2\hspace{5pt}\Rightarrow\hspace{5pt}2a+2b=0\hspace{5pt}\Rightarrow\hspace{5pt}a+b=0 \end{array}$$ From the first option you get three such numbers: $027,117,207$.
From the second option you get two such numbers: $018,108$.
From the third option you get one such number: $009$.
Do the same analysis for the case $b=9$. It is much easier.
If you check the case when $b=9$, you will see that there are no such numbers. So the sum is $27+117+207+18+108+9=486$ if you allow $0$ as a leading digit (i.e. is $009$ a three digit number?). If not - then the sum is $117+207+108=432$ – Dennis Gulko Dec 2 '12 at 12:14