Question

Determine the number of exponents of 8 in 224!.

1. 72
2. 73
3. 74
4. 75​

Answer 1

Given factorial: $224!$

To find: the number of exponent of 8 in the given factorial.

Solution:

Express 8 in terms of its prime factors.

$\[\begin{array}{l}8 = 2 \times 2 \times 2\\ \Rightarrow 8 = {2^3}\end{array}\]$

Find the maximum power of 2 in 224!.

$\[ \Rightarrow \left[ {\frac{{Maximum\,power\,of\,2\,in\,224!}}{3}} \right]\]$

$\[ \Rightarrow \left[ {\frac{{\frac{{224}}{2} + \frac{{224}}{{{2^2}}} + \frac{{224}}{{{2^3}}} + \frac{{224}}{{{2^4}}} + \frac{{224}}{{{2^5}}} + \frac{{224}}{{{2^6}}} + \frac{{224}}{{{2^7}}} + \frac{{224}}{{{2^8}}}}}{3}} \right]\]$

$\[\begin{array}{l} \Rightarrow \left[ {112 + 56 + 28 + 14 + 7 + 3 + 1 + 0} \right]\\ \Rightarrow 221\end{array}\]$

Hence, the number exponents of 8 in 224! is 221.

Answer 2

Number of exponents of 8 in 224! are 73

Given:

• 224!

To Find:

• number of exponents of 8 in 224!

Solution:

• Number of exponent of  a prime number x in   n!  is given by
• ∑  [(n/xᵃ)]   where a = 1 to ∞   where [ ] is greatest integer function

Step 1:

Prime factorize 8

8 = 2 x 2  2  = 2³

Step 2:

Find number of Exponents of  2 in  224!

∑  [(224/2ᵃ)]   where a = 1 to ∞

= [224/2]  +  [224/2²]  + [ 224/2³] + [224/2⁴] + ...

= 112 + 56 + 28 + 14 + 7 + 3 + 1 + 0 + ..

= 221

2²²¹ = 2²¹⁹x2² = (2³)⁷³ x 4 = 4 x 8⁷³

Hence number of exponents of 8 in 224! are 73

Correct option is 2) 73