Q160 / 270 · NDA 2024 (Phase 1) · Mathematics

If $26!={ }_n 8^k$, where $k$ and $n$ are positive integers, then what is the maximum value of $k$ ?

  • A
    6
  • Correct answer
    B
    7
  • C
    8
  • D
    9

Solution

$\because 26!=n 8^k=n 2^{3 k}$ Maximum power of 2 $\begin{aligned} & =\left[\frac{26}{2}\right]+\left[\frac{26}{4}\right]+\left[\frac{26}{8}\right]+\left[\frac{26}{16}\right] \\ & =13+6+3+1=23 \end{aligned}$ But power of $2=3 k$ $\therefore$ Maximum value of $3 k=21$ Hence, the maximum value of $k=7$.

Hence, the correct answer is option: 'B'.

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