Question

The maximum value of p such that 3^p divides 99 × 97 × 95 × … × 51 is –

• A. 11
• B. 14
• C. 13
• D. 12

Answer 1

Answer: (B)

14

Answer 2

99 × 97 × 95 × … × 51

= $\frac{100!}{(100 \times 98 \times 96 \times ... \times 52)}$ × $\frac{1}{50!}$

= $\frac{100! \times 25!}{2^{25} \times 50! \times 50!}$

maximum power of 3 in 100 !

= $\left\lbrack \frac{100}{3} \right\rbrack + \left\lbrack \frac{100}{9} \right\rbrack + \left\lbrack \frac{100}{27} \right\rbrack + \left\lbrack \frac{100}{81} \right\rbrack$

= 33 + 11 + 3 + 1 = 48

maximum power of 3 in 50 !

= $\left\lbrack \frac{50}{3} \right\rbrack + \left\lbrack \frac{50}{9} \right\rbrack + \left\lbrack \frac{50}{27} \right\rbrack$

= 16 + 5 + 1 = 22

maximum power of 3 in 25 !

= $\left\lbrack \frac{25}{3} \right\rbrack + \left\lbrack \frac{25}{9} \right\rbrack$

= 8 + 2 = 10

\ exponent of 3 = 48 + 10 – (22 × 2) = 14