How many 3-term GPs with single-digit terms?

Count ordered triplets \((a,b,c)\) with each in \(\{1,2,\dots,9\}\) such that \((a,b,c)\) forms a geometric progression. Answer: 17.

Concept snapshot

A 3-term GP is \((a,ar,ar^2)\). We need \(a,ar,ar^2 \in \{1,\dots,9\}\) and all integers. So only those rational ratios \(r\) that keep the triple inside \([1,9]\) are valid.

Systematic enumeration by common ratio \(r\)

1) \(r=1\) (constant triples)

All \((k,k,k)\) for \(k=1,\dots,9\) ⇒ 9 GPs.

2) Integer \(r\ge 2\)

\(r=2\): \((1,2,4),(2,4,8)\) ⇒ 2 GPs. \(r=3\): \((1,3,9)\) ⇒ 1 GP. \(r\ge 4\) fails since \(ar^2\ge 16\).

3) Reciprocals \(r\in\{\tfrac12,\tfrac13\}\)

\(\tfrac12\): \((4,2,1),(8,4,2)\) ⇒ 2 GPs. \(\tfrac13\): \((9,3,1)\) ⇒ 1 GP.

4) Proper fractions with integer terms

\(r=\tfrac{3}{2}\): \((4,6,9)\) ⇒ 1 GP. \(r=\tfrac{2}{3}\): \((9,6,4)\) ⇒ 1 GP. Other fractions push terms outside 1–9 or non-integers.

Total (1–9 only): \(9 + (2+1) + (2+1) + (1+1) = \boxed{17}\).

Quick practice

Prove that if \(A,B,C\in\{1,\dots,9\}\) form a 3-term GP with integer ratio, then \(r\in\{1,2,3\}\) or \(r\in\{\tfrac12,\tfrac13\}\).
Show that the only non-integer ratios that still keep terms integral in 1–9 are \(\tfrac{3}{2}\) and \(\tfrac{2}{3}\).
If digits 0–9 are allowed but leading term must be nonzero, how many 3-term GPs exist? (Ans: still \(17\))

GPs with single-digit terms