Question

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

• A. 11
• B. 14
• C. 13
• D. 12

Answer 1

Answer: (B)

$99\times 97\times 95\times \dots \times 51$
$=\frac{100!}{(100\times 98\times 96\times \dots \times 52)}\times \frac{1}{50!}$
$=\frac{100!\times 25!}{2^{25}\times 50!\times 50!}$
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 $3=48+10-(22\times 2)$
$=14$