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A typical book can be printed with 10 6 zeros (around 400 pages with 50 lines per page and 50 zeros per line). Therefore, it requires 10 94 such books to print all the zeros of a googolplex (that is, printing a googol zeros). [4] If each book had a mass of 100 grams, all of them would have a total mass of 10 93 kilograms.
It is a ratio in the order of about 10 80 to 10 90, or at most one ten-billionth of a googol (0.00000001% of a googol). Carl Sagan pointed out that the total number of elementary particles in the universe is around 10 80 (the Eddington number ) and that if the whole universe were packed with neutrons so that there would be no empty space ...
The name of a number 10 3n+3, where n is greater than or equal to 1000, is formed by concatenating the names of the numbers of the form 10 3m+3, where m represents each group of comma-separated digits of n, with each but the last "-illion" trimmed to "-illi-", or, in the case of m = 0, either "-nilli-" or "-nillion". [17]
Gargoogol has 200 Zeros 77.100.228.242 19:35, 7 May 2024 (UTC) []. This number is documented on several wikis such as Fandom, but as user-contributed content it may not be a sufficiently reliable source to support inclusion in the article.
Both are wrong. A googolplex is actually a googol times a googol. That means its equal to 10 to the power of 200 or 1 followed by 200 zeros. Look it up. Skittles 16:44, 28 October 2008 (UTC) 10 to the power of googol and 1 followed by googol zeroes is the same thing. 10^1 = 1 followed by 1 zero. 10^2 is 1 followed by 2 zeroes, etc.
⋮ g 1 = n th tower: 3↑3↑3↑3↑3↑3↑3↑...↑3 (number of 3s is given by the n − 1 th tower) where the number of 3s in each successive tower is given by the tower just before it. The result of calculating the third tower is the value of n, the number of towers for g 1.
Upper bounds on Skewes's number Year near x # of complex zeros used by 2000: 1.39822 × 10 316: 10 6: Bays and Hudson 2010: 1.39801 × 10 316: 10 7: Chao and Plymen 2010: 1.397166 × 10 316: 2.2 × 10 7: Saouter and Demichel 2011: 1.397162 × 10 316: 2.0 × 10 11: Stoll and Demichel
In particular, he proved that for any given numbers ε, ε 1 satisfying the conditions 0 < ε, ε 1 < 1 almost all intervals (T, T + H] for H ≥ exp[(ln T) ε] contain at least H (ln T) 1 −ε 1 zeros of the function ζ(1/2 + it). This estimate is quite close to the conditional result that follows from the Riemann hypothesis.