Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $$A+2B+3C=N $$ where $N$ is a given positive integer.

$A ,B,C\in\mathbb{N}$ vary from $0$ to $\infty$.

How many solutions will be there to this equation?

share|improve this question
1  
Please try and format your questions using $\LaTeX$ notation, and avoid non-standard abbreviations. –  Andreas Caranti Mar 19 '13 at 16:51
    
$N+3-1\choose{3-1}$ , it isn't possible I suppose? –  Inceptio Mar 19 '13 at 17:00
    
Are $A,B,C$ also integers? –  Berci Mar 19 '13 at 17:00
    
Yes A,B,C are intergers –  user1907531 Mar 19 '13 at 17:01
    
@Berci ,ABC are integrs –  user1907531 Mar 19 '13 at 17:04

3 Answers 3

It appears to be sequence A001399 at the Online Encyclopedia of Integer Sequences, where the simple formula $${\rm round}{(n+3)^2\over12}$$ is given, along with lots of interpretations, links, references, and so on.

share|improve this answer

Let $X=C, Y=B+C, Z=A+B+C$, then this is equivalent to finding the number of non-negative integer solutions to $X+Y+Z=N$ s.t. $X \leq Y \leq Z$.

First ignore the restriction on $X, Y, Z$. In total we have $\binom{N+2}{2}$ solutions.

The number of solutions with $Y=Z$ is just the number of non-negative integer solutions to $2Y \leq N$, giving $\lfloor \frac{N}{2} \rfloor+1$. By symmetry we get the same number of solutions for $X=Y, X=Z$.

The number of solutions for $X=Y=Z$ is 1 if $3 \mid N$ and 0 otherwise. We can represent this as $1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil$.

The number of solutions with exactly 2 variables equal is thus $3[(\lfloor \frac{N}{2} \rfloor+1)-(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)]$.

Now going back to the problem, each solution where $X<Y<Z$ corresponds to 6 solutions where $X, Y, Z$ are distinct. Each solution with $X \leq Y \leq Z$ and exactly 2 of them equal corresponds to 3 solutions where exactly 2 of them equal but without the condition $X \leq Y \leq Z$.

We thus get:

\begin{align} & \frac{\binom{N+2}{2}-3(\lfloor \frac{N}{2} \rfloor+1)+2(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{6} \\ & +\frac{3(\lfloor \frac{N}{2} \rfloor+1)-3(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{3} +(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil) \\ & =\frac{\binom{N+2}{2}+3(\lfloor \frac{N}{2} \rfloor+1)+2(1-\lceil \frac{N}{3}-\lfloor \frac{N}{3} \rfloor \rceil)}{6} \end{align}

share|improve this answer
    
This is what I was tryna do(Use greatest int func).:). +1 Great solution. –  Inceptio Mar 20 '13 at 5:10

The result will be equal to the coeeficient of $x^{n}$ in $(x+x^{2}+x^{3}+\dots)(x^{2}+x^{4}+\dots)(x^{3}+x^{6}+\dots)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.