Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
The highest power of 9 dividing 99! completely is
(a)11
(b)20
(c)22
(d)24
Answer
Answer (as printed): D
Explanation
Highest power of 3 in 99! $=\left[\frac{99}{3}\right]+\left[\frac{99}{3^{2}}\right]+\left[\frac{99}{3^{3}}\right]+\left[\frac{99}{3^{4}}\right]$ $$\begin{aligned} & =\left[\frac{99}{3}\right]+\left[\frac{99}{9}\right]+\left[\frac{99}{27}\right]+\left[\frac{99}{81}\right] \\ & =33+11+3+1=48 . \end{aligned}$$ Since $9=3^{2}$, so highest power of 9 dividing $99!=\frac{48}{2}=24$.
Explanation as extracted from the printed page; notation may be imperfect.