My First Problem is Back!

Suppose we need to press the red button $a$ times, blue button $b$ times, yellow button $c$ times, and the green button $20-a-b-c$ times. Here, $a$ , $b$ and $c$ are non-negative integers. You can get the following equation $17a+41b+25c+11(20-a-b-c)=386$ which upon simplifying yields $3a+15b+7c=83$ This is a linear Diophantine equation with 3 variables, and this question is actually asking you how many non-negative integer solutions $(a, b, c)$ satisfy $3a+15b+7c=83$ .

To solve this, we move $15b$ to the right to get $3a+7c=83-15b$ Because $a\geqslant 0$ , $c\geqslant 0$ , so $3a+7c=83-15b\geqslant 0$ , thus $0\leqslant b \leqslant 5$ .

Let $83-15b=n$ where $n$ is a non-negative integer, now we have a new linear Diophantine equation $3a+7c=n$ We first observe the L.H.S. of the equation, the unknown with the smallest coefficient is $a$ which is 3. Divide both sides by 3 and we would get $a+2c+\frac{c}{3}=\frac{n}{3}$ Then, we move all fractions to the right and all the integers to the left. We get $a+2c=\frac{n-c}{3}$ Because $a$ and $c$ are integers, therefore $a+2c$ is also an integer, $\frac{n-c}{3}$ must be also an integer. Let $\frac{n-c}{3}=k$ where $k$ is an integer. Then $c=n-3k$ Substitute $c$ in $3a+7c=n$ with $n-3k$ you get $a=7k-2n$ We have generated our 'solutions formula'. $\begin{cases} a=7k-2n \\ c=n-3k \end{cases}$ From $a=7k-2n\geqslant 0$ , $c=n-3k\geqslant 0$ , and $a+c<20$ , we would get $\frac{2n}{7} \leqslant k \leqslant \frac{n}{3},\quad k<\frac{n}{4}+5$ We now have 2 cases:

Case 1: $\frac{n}{3}\geqslant\frac{n}{4}+5$

Then $n\geqslant60\implies83-15b\geqslant60\implies 0\leqslant b\leqslant1$ , and we have $\frac{2n}{7}\leqslant k<\frac{n}{4}+5$

By substituting back $n=83-15b$ you'll get $23-4b+\frac{5-2b}{7}\leqslant k<\frac{3-3b}{4}+25-3b$ Since $k$ is an integer, $23-4b+\left\lceil\frac{5-2b}{7}\right\rceil\leqslant k<\left\lceil\frac{3-3b}{4}\right\rceil+25-3b$ ( $\lceil x\rceil$ is the smallest integer greater or equal to $x$ )

Denote the number of integer solutions $k$ that satisfy this inequality as $S_k(b)$ , then $\begin{aligned}S_k(b)&=\left(\left\lceil\frac{3-3b}{4}\right\rceil+25-3b\right)-\left(23-4b+\left\lceil \frac{5-2b}{7}\right\rceil\right) \\ &=b+2+\left\lceil\frac{3-3b}{4}\right\rceil-\left\lceil\frac{5-2b}{7}\right\rceil\end{aligned}$

Because $0\leqslant b\leqslant1$ , we have $\left\lceil\frac{5-2b}{7}\right\rceil=1$ , and $S_k(0)=0+2+\left\lceil\frac{3}{4}\right\rceil-1=2 \\ S_k(1)=1+2+\left\lceil0\right\rceil-1=2$

Case 2: $\frac{n}{3}<\frac{n}{4}+5$

Then $n<60\implies83-15b<60\implies 2\leqslant b\leqslant5$ , and we have $\frac{2n}{7}\leqslant k\leqslant\frac{n}{3}$ Substitute $n=83-15b$ , $23-4b+\frac{5-2b}{7}\leqslant k\leqslant\frac{2}{3}+27-5b$ Since $k$ is an integer, $23-4b+\left\lceil\frac{5-2b}{7}\right\rceil\leqslant k\leqslant\left\lfloor\frac{2}{3}\right\rfloor+27-5b=27-5b$ ( $\lfloor x \rfloor$ is the greatest integer smaller or equal to $x$ ) $\begin{aligned}\therefore S_k(b)&=(27-5b)-\left(23-4b+\left\lceil\frac{5-2b}{7}\right\rceil\right)+1 \\ &=5-b-\left\lceil\frac{5-2b}{7}\right\rceil\end{aligned}$ We have $S_k(2)=2,\quad S_k(3)=2,\quad S_k(4)=1,\quad S_k(5)=0$

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$\displaystyle \sum_{b=0}^{5}{S_k(b)}=9$ , therefore, there are 9 possible positive integer solutions for $k$ , substitute $k$ into our previously generated 'solutions formula' and you would get 9 pairs of solutions for $a$ and $c$ .