Diophantine Algorithm Word Problem 1

Question

A says, "We three have P100 altogether". B says, "Yes, and if you had six times as mich and I had one third as much, we three would have still have P100". C says, "It's not fair. I have less than P30". Who has what?

Answer 1

The problem translates into $$\tag1 A+B+C=100$$ $$\tag26A+\frac 13 B+C=100$$ $$\tag3C<30$$ with $A,B,C\in\Bbb N_0$. Eliminating $C$ from $(1)$ and $(2)$, we find $15A=2B$; in particular $2\mid A$ and $15\mid B$, in other words, there is $D\in\Bbb N_0$ with $A=2D$ and $B=15D$. Then $C=100-17D\ge 0$ implies $D\le 5$, and $C=100-17D<30$ implies $D\ge 5$. Hence $D=5$, $$A=10,\quad B=75, \quad C=15. $$

Answer 2

One has $$A+B+C=6A+\frac B3+C=100\Rightarrow \frac BA=\frac{15}{2}$$ It follows $$(A,B)=(2t,15t)$$ Supposing $A,B,t$ integers one has $$(A,B)\in\{(2,15),(4,30),(6,45),(8,60),(10,75),...\}$$ The only couple compatible with $A+B+C=100$ and $C\lt 30$ is $(10,75)$.

Thus $$\color{red}{(A,B,C)=(10,75,15)}$$