Question

find the sum of the digits of the decimal form of the product 2^(1999).5^(2002)

Answer 1

$2^{1999} * 5^{2002} \\ \\ = (\frac{2^{2000}}{2^1}) * (5^{2000} * 5^2) \\ \\ = (\frac{2^{2000}}{2}) * (5^{2000} * 25) \\ \\ = \frac{2^{2000} * 5^{2000} * 25}{2} = \frac{25 * (2 * 5)^{2000}}{2} = \frac{25 * 10^{2000}}{2}$

10^2000 means a number written as 2000 zeros after 1.

(10^2000) ÷ 2 means the number written as 1999 zeros after 5.

(25 x 10^2000) ÷ 2 is the number written as 1999 zeros after 125.

∴ The sum of digits is 1 + 2 + 5 + (1999 x 0) = 1 + 2 + 5 = 8.

8 is the answer.

Answer 2

Step-by-step explanation:

Given: $2^{1999}$ × $5^{2002}$

To find: The sum of digits of decimal form.

For calculation of the above equation,

⇒ $(\frac{2^{1999}}{2^{1}})$ × $(5^{2000}$ × $5^{2})$

⇒ $(\frac{2^{2000} }{2} )$ × $(5^{2000}$ × $25)$

⇒ $\frac{2^{2000 x 5^{2000} x 25 } }{2}$

⇒ $\frac{25x(2x5)^{2000}}{2}$

⇒ $\frac{25x 10^{2000}}{2}$

We know that $10^{2000}$ is written as $2000$ zeroes after $1$.

We know that $\frac{10^{2000} }{2}$ is written as $1999$ zeroes after $5$.

We know that $(\frac{25x10^{2000}) }{2}$ is written as $1999$ zeroes after $125$.

For calculation of the sum of digits,

⇒ $1+2+5 + (1999x0)$

⇒ $1+2+5=8$

The sum of digits of the decimal form of the product of equation is $8.$