Question

If \(a\) and \(b\) are whole numbers such that \(ab = 121\), find the value of \((a-1)b + 1\).

Hint:

When a product fixes two whole numbers, check all factor pairs. If an expression still varies across pairs, the problem needs an extra condition (like “both are prime/non-unit” or “\(a=b\)”).

Answer 1

Step 1 (List all whole-number factor pairs of 121).
\(121 = 11 \times 11 = 1 \times 121\). So ordered pairs \((a,(b)\) that satisfy \(ab=121\) are: \((1,121),\ (11,11),\ (121,1)\). Step 2 (Simplify the given expression).
\[ (a-1)b + 1 = ab - b + 1. \] Since \(ab=121\), this becomes \[ 121 - b + 1 = 122 - b. \] Step 3 (Evaluate for each admissible pair).
\[ \begin{aligned} (a,(b)=(1,121) & 122 - b = 122 - 121 = 1,
(a,(b)=(11,11) & 122 - b = 122 - 11 = 111,
(a,(b)=(121,1) & 122 - b = 122 - 1 = 121. \end{aligned} \] Hence the value depends on which pair is chosen. Step 4 (Exam convention).
Many exam sets intend the non-unit} factorization of \(121\) (i.e., both factors \(>1\)), which forces \(a=b=11\). Under this common convention, the value is \(111\). \[ {122 - b \ \text{(general)}} \qquad {111 \ \text{if } a=b=11} \]