Consider the equation (40)<sub>x</sub> = (132)<sub>y</sub> in some bases x and y . Then a possible set of values of x and y are
\((40)_x = 4x\) and \((132)_y = y^2 + 3y + 2\), with \(x > 4\) and \(y > 3\).
Check option (D), x = 14 and y = 6: \(4 \times 14 = 56\) and \(36 + 18 + 2 = 56\). They match.
(Other options: x = 8, y = 12 gives 32 ≠ 182. x = 12, y = 8 gives 48 ≠ 90. x = 6, y = 12 gives 24 ≠ 182.)
Answer: (D) 14 and 6
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