Question

the maximum power of 2 in 48 factorial​

Answer 1

Step-by-step explanation:

ur ans dude............✌

Answer 2

2⁴⁶

maximum power of 48! =2⁴⁶

Step-by-step explanation:

Given:

48!

To find:

The maximum power of 2 in 48!

Solution:

• Factorials can be expands as n!= n(n-1) ×(n-2)×(n-3)×(n-4)..2×1
• 48! can be written as

⇒48×47×46×45×44×43×42×41×40×39×38×37×36×35×34×33×32×31×30×29×28×27×26×25×24×23×22×21×20×19×18×17×16×15×14×13×12×11×10×9×8×7×6×5×4×3×2×1.

• It took time for manual multiplication, For this case, we verify the highest number power we have to find is a prime number(2).
• The number (2) is neither a prime nor an even number.
• If p is prime number means then the largest power of p in factorial n is given by
•  ⇒($\frac{n}{p}$)+($\frac{n}{p^2}$)+($\frac{n}{p^3}$)+($\frac{n}{p^4}$)+..

If p is prime the highest power of pᵃ present in the factorial is given by

⇒ (highest power of p in n!)

                  a

Distributing the values in the formula we get, in this n=48 and p=2

 ⇒($\frac{n}{p}$)+($\frac{n}{p^2}$)+($\frac{n}{p^3}$)+($\frac{n}{p^4}$)+($\frac{n}{p^5}$)+..

 ⇒($\frac{48}{2}$)+($\frac{48}{2^2}$)+$\frac{48}{2^3}$)+($\frac{48}{2^4}$)+($\frac{48}{2^5}$)+..

Evaluating the powers we get,

 ⇒($\frac{48}{2}$)+($\frac{48}{4}$)+($\frac{48}{8}$)+($\frac{48}{16}$)+($\frac{48}{32}$)+..

(For ($\frac{48}{32}$)=1.5, we does not evaluate the 2⁶ term because it was 64, which means we get a 0.75

For power value, we need a whole number so we do not evaluate the term after 2⁵ term and consider all corresponding terms as 0.

 ⇒24+12+6+3+1+0

 ⇒46

The maximum power of 2 in 48! is 46

Hence

maximum power of 48! =2⁴⁶.