In the 2022 AMC 10A competition, Problem #17 links repeating decimal fractions to integer equations:

Contest Problem Statement:
"How many three-digit positive integers $abc$ are there whose nonzero digits $a, b,$ and $c$ satisfy $$0.\overline{abc} = \frac{1}{3}(0.\overline{a} + 0.\overline{b} + 0.\overline{c}) \text{ ?"}$$ (The bar indicates digit repetition; thus $0.\overline{abc}$ is the infinite repeating decimal $0.abcabc\dots$)."

From Repeating Decimals to Fractions

Recall the fraction formula for repeating decimals:

• A single-digit repeating decimal has denominator $9$: $$0.\overline{a} = \frac{a}{9}, \quad 0.\overline{b} = \frac{b}{9}, \quad 0.\overline{c} = \frac{c}{9}$$
• Their sum divided by $3$ is: $$\frac{1}{3}\left(\frac{a}{9} + \frac{b}{9} + \frac{c}{9}\right) = \frac{a + b + c}{27}$$
• A three-digit repeating block has denominator $999$: $$0.\overline{abc} = \frac{100a + 10b + c}{999}$$

Equating the two expressions: $$\frac{100a + 10b + c}{999} = \frac{a + b + c}{27}$$ Since $999 = 27 \times 37$, multiplying both sides by $999$ yields: $$100a + 10b + c = 37(a + b + c)$$ Expanding and collecting terms: $$100a + 10b + c = 37a + 37b + 37c$$ $$63a = 27b + 36c$$ Dividing both sides by $9$ gives a clean linear Diophantine equation: $$7a = 3b + 4c$$ where $a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are nonzero digits.

Finding the Solutions with Prolog

Notice that if $a = b = c$, then $7a = 3a + 4a = 7a$ holds automatically for all $9$ repeating numbers: $111, 222, \dots, 999$.
Are there any other solutions where the digits are not all equal?

Let's let Prolog search all digit combinations and find every single solution!

Prolog generates all candidate digit triples, applies the arithmetic relation, and enumerates all 13 valid three-digit integers.