Question

The maximum value of $P$ such that $3^{P}$ divides $99 \times 97 \times$ $95 \times \ldots \times 51$ is

• A. 11
• B. 14
• C. 13
• D. 12

Answer 1

Answer: (B)

$ 99 \times 97 \times 95 \times \ldots \times 51 $
$= \frac{100 !}{(100 \times 98 \times 96 \times \ldots \times 52)} \times \frac{1}{50 !}$
$=\frac{100 ! \times 25 !}{2^{25} \times 50 ! \times 50 !} $
maximum power of $3$ in $100 !$
$=\left[\frac{100}{3}\right]+\left[\frac{100}{9}\right]+\left[\frac{100}{27}\right]+\left[\frac{100}{81}\right] $
$=33+11+3+1=48$
maximum power of $3$ in $50 !$
$=\left[\frac{50}{3}\right]+\left[\frac{50}{9}\right]+\left[\frac{50}{27}\right] $
$=16+5+1=22$
maximum power of $3$ in $25 !$
$=\left[\frac{25}{3}\right]+\left[\frac{25}{9}\right] $
$=8+2=10$
$\therefore $ Exponent of $3=48+10-(22 \times 2)$
$=14$