What is the greatest number that can be written with the digits two, five, eight

Mathematics

2 answers:

vodka [1.7K]2 years ago

6 0

Using each of those numbers as digits of a number then it is 852

However, if we raise each number to a power then 2^(5^8) where

5^8 = 390,625 So, 2^390,625 can be estimated by

multiplying the log of 2 (0.30102999566) by 390,625 which equals

117,589.842054688 So, 117,589 is our exponent and then we need to look up the anti-log of .842054688 which equals 6.9511184318 and our number is 6.9511184318 x 10^117,589

DON'T read my comments. I can't delete them.

ivanzaharov [21]2 years ago

5 0

$2^{5^{8}}$

that is a number with more than 100,000 digits

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6 0

2 years ago

Read 2 more answers

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$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
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a_1=4\\
r=6
\end{cases}
\\\\\\
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