Find the number of positive integral solution to this expression

Question

$$ a+\frac{b}{2}+\frac{c}{3}=7 \left(1+\frac{1}{2}+\frac{1}{3} \right) $$ Find the number of positive integral solution.

Answer 1

Find the number of positive integral solution: $a+\frac{b}{2}+\frac{c}{3}=7 \left(1+\frac{1}{2}+\frac{1}{3} \right).$

As noted in the comments, it is equivalent to: $6a + 3b + 2c = 77$.

Consider mod 3 and mod 2: $$\begin{align}2c\equiv2 \pmod 3 \Rightarrow &c\equiv 1 \pmod 3 \Rightarrow c=3m+1;\\ &b\equiv1 \pmod 2 \Rightarrow b=2n+1.\end{align}$$ Substitute these into the original equation: $$6a+3(2n+1)+2(3m+1)=77 \Rightarrow 6(a+n+m)=72 \Rightarrow a=12-m-n.$$ Hence: $$\begin{cases}a=12-m-n>0\\b=2n+1>0\\ c=3m+1>0\end{cases} \Rightarrow \begin{cases}m+n<12\\n>-\frac12\\ m>-\frac13\end{cases} \Rightarrow \sum_{k=1}^{12}k=78.$$ Indeed, when $m=0 \Rightarrow 0\le n\le 11 \Rightarrow 12$.

Here are the solutions: $$\begin{align}(a,b,c)&=(12,1,1) \Rightarrow 1;\\ &=(11,1,4),(11,3,1) \Rightarrow 2;\\ &=(10,1,7),(10,3,4), (10,5,1) \Rightarrow 3;\\ &\ \ \ \vdots \\ &=(1,1,34),(1,3,31),\cdots , (1,23,1) \Rightarrow 12 \\ \\ &\text{So, it is:} \ 1+2+3+\cdots +12=78.\end{align}$$

Answer 2

We can solve this directly.We have $$11\le 6a+3b+2c=77\Rightarrow 1\le a\le 12$$

$$a=n\Rightarrow 3b+2c=77-6n=\text{ odd }$$

It follows that the number of possible values for $b$ is equal to the number of odds from $1$ to $\lfloor {\frac{77-6n}{3}}\rfloor$.

Consequently there are $$1+2+3+\cdots 11=\frac{11\cdot 12}{2}=66\space\text{solutions}.$$