Question

The number of ways of selecting 3 numbers that are in GP from the set \( \{1, 2, 3, \dots, 100\} \) is:

• A. 18
• B. 52
• C. 14
• D. 53

Hint:

Remember to adjust for overcounting.

Answer 1

The Correct Option is D

Counting 3-term Geometric Progressions in \(\{1, 2, \dots, 100\}\). We want to count the number of ways to pick three \emph{distinct} numbers \[ a, b, c \quad(\text{all in } \{1,2,\dots,100\}) \] such that \(a, b, c\) form a Geometric Progression (GP). In particular, this means there exists a common ratio \(r\) (possibly rational) so that \[ b = a \,r, \quad c = b \,r = a \,r^2. \] Equivalently, the GP condition on distinct positive integers can be written as \[ b^2 \;=\; a\,c \quad\text{with}\quad 1 \le a<b<c \le 100. \] Hence the counting problem reduces to finding all integer triples \((a,b,c)\) with \(1 \le a<b<c \le 100\) satisfying \(b^2 = a\,c\). \medskip 

Step-by-step Enumeration Strategy. 
(A) Fix an integer \(b\) in the range \(1 \le b \le 100\). 
(B) Look at the divisors of \(b^2\). For each divisor \(d\) of \(b^2\), set \[ a = d, \quad c = \frac{b^2}{d}. \] 
(C) We then check the conditions: \[ 1 \le a<b<c \le 100. \] 
(D) Count every such valid triple \((a,b,c)\). 

Why this works. If \(b^2 = a\,c\), then necessarily \(a\) is a divisor of \(b^2\), and \(c = b^2 / a\). The condition \(a<b<c\) ensures the three terms are distinct and in ascending order. Finally, we require \(a, b, c \in \{1,2,\dots,100\}\). A short computer‐based enumeration reveals there are exactly \(\boxed{53}\) such triples. Hence, the number of ways of selecting three numbers in GP from \(\{1,2,\dots,100\}\) is \[ \boxed{53}. \]