In the 1967 MAA High School Mathematics Contest (Problem #24), students were challenged to solve this algebra problem:

"The number of solution-pairs in positive integers of the equation $3x + 5y = 501$ is:"
(A) 33    (B) 34    (C) 35    (D) 100    (E) none of these

An algebraic equation where the solutions must be integers is called a Diophantine equation, named after the Greek mathematician Diophantus of Alexandria (c. 250 AD).

To solve $3x + 5y = 501$ algebraically, solve for $x$: $$3x = 501 - 5y = 3(167) - 5y$$ $$x = 167 - \frac{5y}{3}$$ For $x$ to be an integer, the term $\frac{5y}{3}$ must be an integer. Since 3 does not divide 5, $y$ must be a multiple of 3: $$y = 3k \quad (k \in \mathbb{Z})$$ Substituting $y = 3k$ back into the formula gives: $$x = 167 - 5k$$ Because the problem specifies positive integers ($x > 0$ and $y > 0$):

1. $y > 0 \implies 3k > 0 \implies k \ge 1$
2. $x > 0 \implies 167 - 5k > 0 \implies 5k < 167 \implies k \le \lfloor 167 / 5 \rfloor = 33$

So $k$ can be any integer from $1$ to $33$. There are exactly 33 positive integer solution pairs!

In geometry, each integer pair $(x, y)$ corresponds to a lattice point lying on the line $y = -\frac{3}{5}x + \frac{501}{5}$. In this lesson, we write code to search for, count, and display these solutions.

Move the mouse over a dotted box for more information.

• Lattice points: Points whose coordinates $(x, y)$ are both integers are called lattice points. The equation $3x + 5y = 501$ describes a straight line with slope $-3/5$. Stepping from one solution to the next means decreasing $x$ by 5 and increasing $y$ by 3!
• Smallest and largest solutions:
  • Largest $x$: $k = 1 \implies x = 162, y = 3$
  • Smallest $x$: $k = 33 \implies x = 2, y = 99$