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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?

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    SORRY I am new and i will edit this2012-11-29

1 Answers 1

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I'm answering your question just because your'e new. Please format your questions better in the future.
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.
This way you can find all such numbers and hence their sum.

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    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$2012-12-02