How many digits are in the number 2^2009 × 5^2005?

I've no idea how to begin.

2. Originally Posted by judocallin02

How many digits are in the number 2^2009 × 5^2005?

I've no idea how to begin.
Take logs to base 10: $\displaystyle \log(2^{2009}\times5^{2005}) = 2009\log2 + 2005\log5$. My calculator gives that as 2006.2041, which means that $\displaystyle 2^{2009} \times 5^{2005}$ lies between $\displaystyle 10^{2006}$ and $\displaystyle 10^{2007}$.

Alternatively, $\displaystyle 2^{2009} \times 5^{2005} = 2^4\times(2\times5)^{2005}$, from which you can estimate the answer.

The second method is much cleaner in this case, but the log-to-base-ten method works more generally.

3. thanks

4. Hello, judocallin02!

I recommend Opalg's second method . . .

How many digits are in the number: .$\displaystyle 2^{2009} \cdot5^{2005}$ ?

$\displaystyle 2^{2009}\cdot5^{2005} \;=\;\left(2^4\cdot2^{2005}\right)\cdot5^{2005}$

. . . . . . . $\displaystyle = \;2^4\cdot\left(2^{2005}\cdot5^{2005}\right)$

. . . . . . . $\displaystyle = \;2^4\cdot(2\cdot5)^{2005}$

. . . . . . . $\displaystyle = \;2^4\cdot10^{2005}$

. . . . . . . $\displaystyle = \;16\cdot\underbrace{1000\hdots000 }_{\text{2005 zeros}}$

. . . . . . . $\displaystyle =\;16\underbrace{000\hdots000 }_{\text{2005 zeros}}$

Therefore, the product has $\displaystyle {\color{blue}2007}$ digits.