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

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

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

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

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

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

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

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

$x \text { has exactly } 361 + 1 = 362 \text { digits.}$ Tags amount, calculator, digits Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post CoarxFlow Math Software 5 August 26th, 2016 09:37 PM Chikis Elementary Math 11 October 22nd, 2013 05:55 PM gelatine1 Algebra 10 July 15th, 2013 07:00 PM CarpeDiem Elementary Math 7 July 13th, 2012 08:35 PM autumnsandsprings Algebra 2 May 11th, 2012 03:26 AM

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