Aspire Faculty ID #15326 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

If $(123)_b = 291$, then the value of the base $b$ is …

Solution

$(123)_b = 1b^2 + 2b + 3 = 291$. $\Rightarrow b^2 + 2b + 3 = 291 \Rightarrow b^2 + 2b - 288 = 0$. Solving: $b = 16$ or $b = -18$. Base must be positive → $b = 16$.
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