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April 12th, 2018, 09:13 AM   #11
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Quote:
Originally Posted by Student2018 View Post
So there are 1 + 361 = 362 digits?
Yes

$1 \le a < 10,\ a \in \mathbb R,\ n,\ x \in \mathbb Z,\ n \ge 0, \text { and } x = a * 10^n \implies$

$\text { there are exactly } n + 1 \text { digits in } x.$

$2.03 * 10^{90} < 2^{300} < 2.04 * 10^{90} \implies$

$(2.03 * 10^{90})^4 < (2^{300})^4 < (2.04 * 10^{90})^4 \implies$

$2.03^4 * 10^{360} < 2^{1200} < 2.04^4 * 10^{360}) \implies$

$ 16.98 * 10^{360} < 2^{1200} < 17.32 * 10^{360}) \implies$

$2^{1200} \approx 17 * 10^{360} = 1.7 * 10^{361} \implies$

$x \text { has exactly } 361 + 1 = 362 \text { digits.}$
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