Notable Properties of Specific Numbers
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An upper bound on the processing rate embodied by the human brain, based on 10^{11} neurons with 10000 dentrites each and a firing rate of 1000 per second. A bruteforce simulation of this could be done with about 10^{18} floatingpoint operations per second, although in practice an optimised simulation would probably be used instead, reducing that number to about 10^{14}. See also 10^{9}.
A repunit prime with 19 digits. See 2446 for more.
An episode of The IT Crowd (season 1, "Calamity Jen") featured a parody of the British emergency telephone dispatch system, in the form of an advertisement claiming that the existing number 999 would be replaced by "0118 999 88199 911 9725 3", and featured a catchy jingle to help viewers remember the number. See also 525600, 8675309, 10^{12}, 10^{27} and 10^{1010}.
6.2415090744×10^{18} = 1/1.602176634e19
The number of electrons that carry a coulomb of charge, by the new (2019) definition, see 1.602176634×10^{19}.
A very rough estimate of the number of grains of sand on all of Earth's beaches. This is based on an estimated 50,000 km of beach coastline with average beach width of 30m and depth of 5m, with sand grains that are 1 mm^{3} in size. See also 3×10^{23} and 5.6×10^{24}.
8315553613086720000 ≈ 8.315553...×10^{18}
The number of symmetries of the Leech lattice.
9223372036854775807 = 2^{63}1 ≈ 9.22337×10^{18}
This is the largest positive integer expressible in a 64bit signed 2's complement representation. See also 9007199254740992 and 1.797693134862×10^{308}.
An estimate of the total number of insects living in the world.^{61}
See also 6771000000, 75250000000, 10^{14} and 5×10^{30}.
12157692622039623539 ≈ 1.216×10^{19}
12157692622039623539 = 1^{1} + 2^{2} + 1^{3} + 5^{4} + ... + 3^{19} + 9^{20}, the sum of its own digits raised to consecutive powers. See 135.
Approximate value of the Planck mass in GeV.
See also 10^{101056}
12345678910987654321 ≈ 1.23456...×10^{19}
A prime number that is easy to remember; see 2446 for more. See also 2.959364...×10^{21077}.
13043817825332782212.349... ≈ 1.304381...×10^{19}
The square root of 2^{127}. See 1597463007.
An integer approximation to the square root of 2^{127} better known by the hexadecimal form of its floatingpoint representation, 0x5f3759df; see that entry for more.
18446744073709551615 = 2^{64}  1 ≈ 1.8447×10^{19}
This number has two legends associated with it.
The first legend is ancient and concerns the origin of the game of chess. King Shirham of India offered the inventor of the game any reward he cared to name, and the inventor asked for wheat: 1 grain for the first square on the chessboard, 2 grains for the second, 4 for the third, 8 for the fourth and so on — doubling each time. The king was surprised, thinking this request was almost insultingly modest (thinking it would only add up to a few bushels), but obliged and instructed his servants to measure out the inventor's reward. They spent some time trying to do this, and eventually realised that all the wheat in all the kingdoms in the world would not come close to the amount specified. The number of grains of wheat requested is 2^{64}  1. (That is over 400 times the world's total wheat production in the year 2003, assuming a grain of wheat is 20 cubic millimeters.)
The second legend was created by Edward Lucas, the inventor of the Tower of Hanoi game, in 1884. As this story goes, at the centre of the world (or, in another version, at the top of a mountain in the Himalayas), is a set of three rods bearing golden discs. There are 64 discs, and they are all different sizes. When the world was created, they were stacked on one of the rods, with the largest at the bottom and the smallest at the top. Ever since then, the priests have worked continually transferring the discs from one rod to another. They can only move one disc at a time, and they are not allowed ever to put a larger disc on top of a smaller disc. When they succeed in stacking all 64 discs on the second rod in a single stack as they were when the world was created, the world will end. The number of steps required to carry out this process is 2^{64}  1.
The fact that the number comes up in these two different stories seems even more of a coincidence when you consider that in the chess legend, the number is the sum 1+2+2^{2}+2^{3}+...+2^{62}+2^{63}, whereas in the Lucas legend it is 2^{64}1. These happen to be the same sum, but arrived at in two different ways.
See also 2.3148855308184×10^{16}, 10^{137}, and The Lynz.
18446744073709551616 = 2^{64} ≈ 1.8447×10^{19}
This is 2^{64} = 2^{26} = (((((2^{2})^{2})^{2})^{2})^{2})^{2}.
There are two ways to define an operator that follows next after addition, multiplication and exponents. Both definitions specify a function that takes two numbers A and B, and whose value is
A ^ A ^ A ^ ... ^ A
where the number of A's is B. But there are two obvious ways to define this, depending on whether you evaluate it from right to left or from left to right:
((..((A^A)^A)^...)^A)^A or A^(A^(A^...^(A^A)..))
If you group the parentheses from the right, you get the hyper4 operator, which is generally accepted as a better definition. If you group the parentheses from the left, you get something that grows much more slowly, but still pretty fast, and which I studied quite a bit in high school (late 1970's). I used the somewhat confusing name "powerlog" to refer to the function, as in "2 powerlog 7 equals 2^{64}". If you denote the operation by "_{④}", then
A_{④}B = (((A ^ A) ^ A) ^ ... ) ^ A = A^{A(B1)}
One reason the hyper4 definition is chosen over this one is that hyper4 cannot be reduced to a simple formula like this. Like each of the functions before it, hyper4 cannot be defined as a simple finite combination of other "lower" functions.
There is more on these operators on my hyper4 page and in this section of my large numbers page.
I have been asked several times whether the hyper4 operator can be extended to the reals (to compute, for example, π^{④}e). This issue is discussed at length here.
As discussed above, the lowervalued hyper4 operator can be expressed in terms of a double exponent: A_{④}B = A^{A(B1)} — so it is easily extended to reals and complex numbers. For example, see 4979.003621.
(2^{64} also happens to be the largest integer I have ever memorised, except for trivial ones like vigintillion.)
43252003274489856000 = 8!×3^{7} × 12!×2^{11} / 2 ≈ 4.3252×10^{19}
(Rubik's Cube)
The number of ways to arrange a 3×3×3 Rubik's Cube. The centre cubelets are assumed to be stationary. The 8 corners can be arranged into any of the 8!=40320 possible positions, and all but one can be rotated into any of 3 different rotations (the total rotation of all 8 pieces always adds up to 360^{o}); this gives a factor of 8!×3^{7} for the corner pieces. There are 12 edge pieces, which can be put in 12!=479001600 possible positions, and there are 2^{11} rotation combinations (due to another rotation constraint). An additional factor of 2 is lost because if you want to swap a single pair of corners, you must either have a third corner out of place, or swap a single pair of edges. Alternately, if any two edges are out of place, a third edge or a pair of corners must be out of place.
Using the Thistlethwaite cubesolving algorithm from 1981 [147], the cube can be solved in progressive stages that gradually reduce the remaining possibilities through a chain of subgroups (a nested hierarchy of progressivelysmaller subsets of positions) of orders 21119142223872000, 19508428800, and 3674160, thence to the solved state. For many years this method, and optimisations thereof by later authors, held the record for the best known solution algorithm (measured in terms of the number of moves required in the worst case: 52 in Thistlethwaite's original, down as low as 29 in the method of Mike Reid). In 2010 it was proven that all positions of the Cube can be solved in no more than 20 moves (in which halfturns count as a single move). In 2014 a similar result of 26 quarterturns was proven.
Over Three Billion Combinations
This number became a rather wellknown example of "innumeracy", when the packaging of the puzzle (as sold in the United States by Ideal Toy Corporation) said that the puzzle had "over 3 billion combinations". While not technically wrong, it is quite an understatement to say the least — the actual number of combinations is greater than the square of 3 billion. Apparently, the company thought that customers would not understand "over 43 quintillion" quite as well as "over 3 billion".
If the rotation and swap constraints are avoided by physically disassembling the cube, you get a number 12 times as large, 519024039293878272000=8!×3^{8}×12!×2^{12}.
Numberphile has a related video, 43,252,003,274,489,856,000 Rubik's Cube Combinations.
See also 3674160, 7.4012×10^{45}, 2.8287×10^{74}, 1.5715×10^{116}, and 1.9501×10^{160}.
The product of the gravitational constant (approximately 6.67384(80)×10^{11}) and the mass of the Sun (approximately 1.98892(25)×10^{30}) and used in Solar System dynamics calculations (such as when generating ephemeris parameters for predicting orbital trajectories of spacecraft). The product can be computed far more accurately than either of the terms.
A realistic answer to the "number noone else has ever thought of" question discussed in the entry for 1.76×10^{67}. It is 100 times 117000000000 times 12833424. See also 3112066128.
This is the smallest 6perfect number. Its divisors add up to exactly 6 times itself. Its prime factorisation is: 2^{15}×3^{5}×5^{2}×7^{2}×11×13×17×19×31×43×257. This is the smallest case of a Kperfect number where K itself is a perfect number (:
158962555217826360000 ≈ 1.59×10^{20}
This is cited by Numberphile as being the number of ways to set up a Nazi army Enigma machine, the type with three rotors (selected out of 5 provided), each with 26 possible starting positions, and a plugboard on which ten pairs of letters are connected with patch cables. As explained in the video, the number of combinations works out to C(3,5) × 26^{3} × C(20,26)/(10!×2^{10}) = 5×4×3 × 26^{3} × 26!/(6!×10!×2^{10}) = 158962555217826360000.
295147905179352825856 = 2^{68} ≈ 2.95×10^{20}
The smallest power of 2 that contains each of the digits (0 through 9) at least once.
See also 2048, 536870912, 1026753849, and 2504730781961.
336365328016955757248 ≈ 3.363653...×10^{20}
This is 6954572^{3}, and is just 1 less than a sum of two cubes: 5444135^{3} + 5593538^{3} = 6954572^{3}  1. It is an example of a nearmiss to Fermat's last theorem, of which other examples include the famous 9^{3}+10^{3} = 12^{3}+1 = 1729, and two larger examples (see 1030301000 and 588522607645608). All are defined by a set of three generating functions on page 82 of Ramanujan's "Lost notebook". See this article for a thorough discussion.
The average (expected value) of the final roll on Homestuck's CATENATIVE DOOMSDAY DICE CASCADER. The Cascader is a parody of D&D dice and gameplay techniques. First, a normal die is rolled to get a number from 1 to 6. If they roll N, the player is given N many additional rolls of Ssided dice, where S is initially equal to N. For each of these additional rolls, the player is rolling an Ssided die, and after each roll S is multiplied by the value that was just rolled. In the linked cartoon N is 6, and S takes on the values: 6, 12, 96, 3360, 9817920, and 50090870753280. Sadly, the player rolled a 1 on the final roll and suffered a critical failure.
In theory S can square each time, so rolling the highest possible value each time gives 6^{32} ≈ 7.958×10^{24}. The average is reduced to 1/6 of this since the 5 squarings are only possible if you get a 6 on the first roll, and by another factor of (1/2)^{5} = 0.015625 because on average each roll will be in the middle of the possible range. It is actually reduced more than that because each roll affects the range of subsequent rolls. This post on Reddit says that the average final outcome is 846557758482794171303705/1728, about 4.899×10^{20}.
519024039293878272000 = 8!×3^{8}×12!×2^{12} ≈ 5.1902×10^{20}
(Rubik's cube by reassembly)
This is the number of combinations of the 3×3×3 Rubik's Cube if you're allowed to take it apart and put it back together. See also 43252003274489856000 and 101097362223624462291180422369532000000.
(a "grillion")
A "grillion" is "ten to the eleventeenth power", according to Urban Dictionary.
See also zillion.
6670903752021072936960 = 2^{20}×3^{8}×5×7×27704267971 ≈ 6.6709×10^{21}
(Sudoku)
This is the number of possible solution patterns to Sudoku puzzles, based on the normal baseline Sudoku requirements (one of each digit in each row, column, and 3×3 subsquare). The number has been calculated and verified by two independent researchers using different exhaustive counting (bruteforce search) algorithms^{33}. This number counts two patterns as distinct even if a "relabeling" (e.g. changing all 3s to 7s and viceversa) would make them equivalent. Many arrangements do not make for very interesting puzzles because (for example) they might be unsolvable unless you supply a lot of numbers in one area of the grid, making for an uneven starting pattern. See also 362880, 3546146300288 and 5524751496156892842531225600.
The value of the Earth (in US dollars) as calculated by reddit user ShadyPotat0, based on the idea of selling all its elements (iron, gold, vanadium, etc. but not diamonds, lumber, etc.) at 2014 market prices. See also 5×10^{15} and 6.874×10^{15}.
(megaparsec)
The length in meters of a "megaparsec", a unit of distance used in astronomy when discussing the size of the entire observable universe and other largescale features, such as the distances of quasars and other very distant objects. See also 5878625373183.6 and 3.08568025×10^{16}.
31858749840007945920321 = 422481^{4} ≈ 3.1859×10^{22}
This is 422481^{4}, equal to 95800^{4} +217519^{4} +414560^{4}. This is the simplest known counterexample to the conjecture of Euler that an n^{th} power cannot be expressed as the sum of less than n smaller n^{th} powers. In this case it's the sum of 3 4^{th} powers. Of course, it can never happen with a sum of two 3^{rd} powers, because that would be a counterexample to Fermat's Last Theorem. See also 216.
An old (pre2004) estimate of the number of stars in the observable universe, based on observations by the Hubble telescope. The current estimate is closer to 3×10^{23}. See also 7.5×10^{18}.
60573000000000000000000 = 6.0573×10^{22}
The extraterrestrial population of the visible universe according to Thomas Dick in his 1840 work Sidereal Heavens and Other Subjects Connected with Astronomy, As Illustrative of the Character of the Deity and of a Infinity of Worlds. See also 21891974404480 and 3×10^{23}.
70000000000000000000000 = 7×10^{22}
This is another old estimate of the number of stars in the observable universe, based on the strip survey estimate of Dr. Simon Driver's team, announced in early 2004. An earlier estimate was 3.2×10^{22}, and the current estimate is closer to 3×10^{23}.
See also 10^{97}.
151115727451828646838272 = 2^{77}
When grouped in threes, the digits of 2^{77} have 6 palindromeic parts: 151,115,727,451,828,646,838,272. See also 134217728.
278914005382139703576000 ≈ 2.789×10^{23}
This is 2^{6}×3^{5}×5^{3}×7^{2}×11×13×17×19×23×29×31×37×41×43×47, and is the smallest number with over a million distinct factors (1032192 to be exact). See also 12, 840, 45360, 720720, 3603600, 245044800, 2054221614063184107682218077003539824552559296000 and 457936×10^{917}.
(Number of stars in the visible (or "observable") universe)
A recent (2010) study by van Dokkum using the Keck telescope estimated that there are 300 sextillion stars in the visible universe. (An earlier estimate from early 2004 was 7×10^{22}). These numbers are really hard to estimate because astronomers cannot see most of the stars in our galaxy (due to clouds of dust, and because most stars are too faint) — nor can they see most of the galaxies (for the same reasons) — so there needs to be much statistical guesswork. Even the definitions of "star" and "visible universe" are subject to debate and change.
By coincidence, 3×10^{23} is about half a mole of stars. It also happens that if you try to take 3×10^{23} things and arrange them along a straight line stretching from our location to the edge of the visible universe, they would be almost exactly a mile apart (see 1609.344 and 4.45×10^{26}). Of course, stars are much bigger than a mile (even those massive enough to have collapsed into a neutron star or stellar black hole) but it gives a sense of the relative size of the universe and how much empty space there is.
See also 10^{97}.
357686312646216567629137 ~= 3.58×10^{23}
Starting with a prime, you can often add a digit to the beginning to make another prime. For example: 7 → 17 → 317 → 6317 → 26317, etc. 357686312646216567629137 is the largest number in base 10 that you can get this way. If 0 is allowed as a digit, then the task of finding the largest has no welldefined solution. (There are many, arbitrarily large, primes similar to 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003 = 10^{101}+3). See also 33333331, 73939133, 381654729, and 3608528850368400786036725.
602214179000000000000000 = 6.02214179(30)×10^{23}
(Avogadro's number)
The current mostaccurate form of the "Avogadro constant" that many of us learned in chemistry class as "6.02×10^{23}". The (30) at the end of "6.02214179(30)" represents the standard uncertainty (error) in the value: It is a more concise way to say "6.02214179 plus or minus 0.00000030".
This is the number of carbon atoms in 12 grams of pure carbon, unbound and in ground state. In general it is very close to the number of atoms in N grams of an element with atomic weight N. (The term atomic weight has since been replaced by "relative atomic mass".) For different substances the correspondence varies slightly, primarily because of binding energy in the nucleus and mass of the electron. (Note that "atomic weight" is not the same as the integer "atomic number", which is hardly ever close enough to be useful.)
A lot of people identify the Avagadro number as the largest number they have had to use, or in many cases, actually memorise.
The amount (in US dollars) demanded in a lawsuit filed in 2009 August by one Dalton Chiscolm against Bank of America. This broke the previous record from 2007, but was surpassed in 2014.
This is notably much bigger than 4^{32}, even though it is made from the same 3 digits. This illustrates the principle of powertowers demonstrated by 2^{341} and 6pt1.86×10^{3148880079}.
3608528850368400786036725 ~= 3.6×10^{24}
This number is polydivisible, meaning that for any N, its first N digits are divisible by N: 3 is divisible by 1, 36 is divisible by 2, 360 is divisible by 3, 3608 is divisible by 4, etc. There is no larger number with this property. See also 73939133, 381654729, and 30000600003.
Another very rough estimate of the number of grains of sand on all of Earth's beaches. This is based on an estimate that there are 7×10^{14} cubic meters of sand, and that a grain of sand is 1/8 of a mm^{3} in volume. See also 7.5×10^{18} and 3×10^{23}.
21655649541169643693329642 ≈ 2.165×10^{25}
A reader wrote to discuss the issue of whether certain welldefined, but really large numbers (like the Moser) can be said to "exist", given that there is no way to write out all their digits, or in some cases (like the GrahamRothschild number), no exact representation in any easily understood notation.
I suggested a rather strict definition of existence, which we can call "actual physical existence":
A number exists only when there is an object that has that many things in it.
So for example, if there is a stone that contains exactly 21655649541169643693329642 atoms, then the number 21655649541169643693329642 can be said to exist, but once that stone gains or loses an atom the number 21655649541169643693329642 doesn't exist anymore.
This definition seems to depend on an observer (kind of like the "if a tree falls in a forest" argument) because if the stone isn't available how can we count the atoms?
Of course, the practical difficulty of counting atoms (see Hill et. al [212]) makes this definition pretty useless anyway. If we do ever manage to count atoms in stones, the same problem would still arise when considering larger numbers like the number of atoms in a planet.
Given that we can't even prove the physical existence of a large number like 21655649541169643693329642, I find it unnecessary to worry much about the existence or nonexistence of larger numbers, particularly when they are rigorously defined.
Mathematicians tend to deal with types of existence that are very abstract, sometimes even too abstract for the philosophers. If they are willing to discuss a number and make definitive claims about it, then I feel that I can too.
For more on this and similar issues, see my discussion of superclasses.
24547284284866560000000000 ≈ 2.45×10^{25}
24547284284866560000000000 has this special property shared with 2592:
2^{4} 5^{4} 7^{2} 8^{4} 2^{8} 4^{8} 6^{6} 5^{6} 0^{0} 0^{0} 0^{0} 0^{0} 0^{0} = 24547284284866560000000000
(using the standard definition of 0^{0}=1). It is known that apart from 2592, there are no others below 10^{100} (see A135385).
The size of the universe in meters.
See also 3×10^{23}.
1000000000000000000000000000 = 10^{27}
(Approximately) the power output of the Sun in watts.
10^{27} is one octillion. Walt Whitman used octillions, as well as several other large number names including decillions, sextillions, quintillions, quadrillions, trillions, ten billion and ten thousand, in his poetry.
See also vigintillion and the song lyric references in 525600, 8675309, 10^{11}, 4.28...×10^{369693099}, and 10^{10000000000}.
10^{27} is the value of the fake (hoax) SI prefix "xera", which was planted on Wikipedia ^{111} and made it all the way to the New York Times ^{112} before anyone noticed.
2235197406895366368301560000 ≈ 2.235197...×10^{27}
If a standard 4suit 52card deck is shuffled (randomly) and the cards are dealt to four players, the odds of all four of them getting a full suit are 1 in 2235197406895366368301560000. This number is (52!)/(13!^{4}×4!), because you first shuffle the deck (52!), then divide the cards into 4 groups of 13, but it doesn't matter what order the 13 cards in each hand are dealt (divide by 13! for each hand), and it also doesn't matter which player gets which suit (divide by 4!).
See also 158753389900, 635013559600 and 8.065817...×10^{67}.
5524751496156892842531225600 ~= 5.5×10^{27}
The number of 9×9 Latin Squares. See also 3546146300288 and 6670903752021072936960.
Approximate mass of the Earth, in grams. See also 1.988435×10^{33}.
7000000000000000000000000000 = 7×10^{27}
(Approximately) the number of atoms in a 70kilogram (154pound) human being. To calculate your number of atoms, multiply your weight in kilograms by 10^{26}. See also 713580 and 6.32×10^{28}.
10888869450418352160768000000 = 27! ≈ 1.0889×10^{28}
Just another gratuitous 27 reference. See also 10^{3.0056206947796095239×1029}.
The nextlarger number after 24310 that occurs at least 5 times in Pascal's triangle [142]. It is 103!/(40!×63!) and is also 104!/(39!×65!); setting these equal and removing all the common terms we get
1/40 = 104/(64×65)
with the significant thing being that these are all products of Fibonacci numbers:
1/(5×8) = 8×13/(5×8×8×13)
The nextsmaller one fitting this pattern is 3003; Singmaster showed [142] that the pattern continues with the nextlarger one being 3.537835...×10^{204} wherein the relevant Fibonacci products are 21×34 = 714, 13×21 = 273, and 21×21 = 441.
Perhaps of more interest, it is known (see comments in OEIS sequence A3015) that there are no numbers between 24310 and 61218182743304701891431482520 that occur more than 4 times as a binomial coefficient. Singmaster [142] found this a bit surprising, and conjectured that perhaps such multiplicities have a fairly low upper bound, (like 8, 10, or 12).
Approximate number of particles (protons, neutrons and electrons) in a 70kilogram (154pound) human being. This is the mass in grams times Avogadro's number times 1.5 (based on the approximation that the number of electrons is half the number of protons + neutrons).
See also 7×10^{27} and 10^{1029}.
77777733332222222222222222222 ≈ 7.777..×10^{28}
This is conjectured to be the largest number, not containing any 1 digits, with a persistence of 11. Except for the first step, its persistence chain is the same as that for 277777788888899.
1000000000000066600000000000001 ≈ 1.0×10^{30}
Called "Belphegor's prime", this number is a palindrome, with a 666 in the middle and 13 zeros on each side. Numberphile has a video on it, "Belphegor's prime". See also 10^{(6.6556570552...×10668)}
Mass of the Sun in kilograms. The "(25)" indicates the estimate of the uncertainty in the number: 1.98892±0.00025. See 10^{57}.
An estimate of the number of prokaryotes (tiny organisms without a nucleus, including bacteria) on the Earth^{123}. Produced by William Whitman and colleagues at the university of Georgia, the estimate includes 26×10^{28} organisms within the top 8 meters of the ground, 12×10^{28} in the water and oceans, about 80×10^{28} on land but below the 8meter point, and 355×10^{28} in the ocean floor. The same scientists measured the mutation rate and estimated that every 20 minutes, somewhere on Earth a new species of bacteria comes into existence. Since prokaryotes comprise the vast majority of all living cells^{6}, 5×10^{30} is also the total population of living things on the planet.
See also 10^{14}.
10331233010526315789473682240000
Erroneous version of Mayan Coba Stela 1 number; see 10331233010526315789473684112000.
10331233010526315789473684112000 = (20^{19}1)×(13/19)×20^{3}×18 ≈ 1.033×10^{31}
The number of days represented in the Mayan date numbering system by the number 13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.0.0.0.0_{20}, which uses base 20 in each position except the second to last (see 5126 and 1872000). There are 20 digits of "13" and four 0's. This appears in an inscription on Stela 1 at Coba in Yucátan (see [213] figure 3, columns M and N; also [155]. The symbols in the top half are all 13's next to glyphs representing units of time each 20 times longer than the next, after that are zeros for the b'ak'tun, k'atun, tun and winal units, then "4 Ahau" and "8 Cumku" among some other glyphs).
The following table illustrates how the expression (20^{19}1)×(13/19)×20^{3}×18 is equal to 13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.0.0.0.0_{20}. (If you don't get the "(13/19)" part, consider the fact that 555 equals (10^{3}1)×(5/9). )

Following the same pattern, twenty 13's and four 0's is (20^{20}1)×(13/19)×20^{3}×18, which works out to 10331233010526315789473684112000.
The value is often given as 10331233010526315789473682240000, but that is 13×20^{3}×18 less, or 13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.13.0.0.0.0.0_{20}, with nineteen 13's and five 0's.
See also 23040000000.
13333398333445333413833354500001 ≈ 1.33334×10^{31}
This is the number of ways to make change for $1,000,000 entirely out of coins, with available coin denominations of 1, 5, 10, 25, 50, and 100 cents (which for a long time have been the denominations of coins in the United States). The multiple sets of 333's in its digits is a pattern that appears in the answer to the same problem for other powersof10 dollar amounts. This is because the number can be expressed as a 5^{th}degree polynomial in x, with x equal to 10^{6} (see 293 for more). The reasons for this are explained in quite some detail by Mathologer in the video making change for a googol dollars, which presents the derivation in the 2005 paper "Changing Money" by Andrew Neitsch.
19252884016114523644357039386451 = 1205967×primorial(61) + 19111438098711663697781258214361 ≈ 1.9252884...×10^{31}
The smallest known set of 7 consecutive primes that are spaced an even distance apart starts with this number. The spacing is 210. This set of primes, called a "CPAP7" (for "continuous primes in arithmetic progression"), was discovered by Laurent Fousse & Paul Zimmermann in 2004^{21}. It is probably not the smallest; these types of primes are much more common at larger sizes, and therefore an exhaustive search for the first one takes longer than a coordinated search for a larger one. It is suspected that the smallest is about 22 or 23 digits long. See also 47, 251, 9843019 and 121174811.
91409924241424243424241924242500 = 1^{10} + 2^{10} + 3^{10} + 4^{10} + ... + 999^{10} + 1000^{10} ≈ 9.1409924...×10^{31}
The sum of the 10^{th} powers of the numbers from 1 to 1000. Notice the pattern in the digits — sort of regular, yet sort of irregular.
Around the year 1690, Jacques Bernoulli was studying the task (which was common at the time) of computing sums of powers. In the process he discovered the Bernoulli numbers, and used his new formulas to quickly calculate this sum as a demonstration of the importance of his discovery.
The Bernoulli numbers appear as coefficients in infinite series for many things including the sums of powers, various other combinatoric formulas, the Gamma function, etc. The first few Bernoulli numbers are: B_{0}=1, B_{1} = ^{1}/_{2}, B_{2}=^{1}/_{6}, B_{3}=0, B_{4}=^{1}/_{30}, B_{5}=0, B_{6}=^{1}/_{42}, B_{7}=0, B_{8}=^{1}/_{30}, B_{9}=0, B_{10}=^{5}/_{66}, B_{11}=0, B_{12}=^{691}/_{2730}, B_{13}=0, B_{14}=^{7}/_{6}, B_{15}=0, B_{16}=^{3617}/_{510}, B_{17}=0, B_{18}=^{43867}/_{798}, B_{19}=0, B_{20}=^{174611}/_{330}, B_{21}=0, B_{22}=^{854513}/_{138}, B_{23}=0, B_{24}=^{236364091}/_{2730}, ... Notice that the odd Bernoulli numbers starting with B_{3} are all zero, and the even ones starting with B_{2} alternate in sign. Also, after a point the size of the numerator starts to grow quickly. To compute the Bernoulli numbers, most mathematicians use the "generating function" method:
f(x) = x / (e^{x}1) = SUM_{n=0...inf}[B_{n} x^{n} / n!]
The way you determine the B_{n} from this is by finding the value of the n^{th} derivative of the function at 0, which is equivalent to calculating the coefficients of the Taylor series for the function:
f(x) = f(0) + f'(0) x / 1! + f''(0) x^{2} / 2! + f'''(0) x^{3} / 3! + ...
Since f(x) and each of its derivatives are undefined for x=0, you actually have to use the limit as x approaches 0. Or, you can use a symbolic math program — for example, in Maxima type taylor(x/(%e^x1),x,0,10); then multiply the n^{th} coefficient by n!
If you don't want to go to all this trouble the Bernoulli numbers can also be calculated by a simple iterative algorithm:
B_{0} = 1
B_{n} =  n! SUM_{k=0..(n1)} [B_{k}/(k! (n+1k)!)]
The sums for the first few B_{n}'s expand out like so:
B_{1} =  1! ( B_{0}/(0!×2!) )
B_{2} =  2! ( B_{0}/(0!×3!) + B_{1}/(1!×2!) )
B_{3} =  3! ( B_{0}/(0!×4!) + B_{1}/(1!×3!) + B_{2}/(2!×2!) )
(etc.)
The factorials get to be rather big pretty quickly. You can avoid such big numbers by using a different, less elegant iterative algorithm, illustrated here:
B_{0} = 1
B_{1} =  B_{0} / 2
B_{2} = (3 B_{1}  B_{0}) / 3
B_{3} = (6 B_{2} + 4 B_{1} + B_{0}) / 4
B_{4} = (10 B_{3}  10 B_{2}  5 B_{1}  B_{0}) / 5
B_{5} = (15 B_{4}  20 B_{3} + 15 B_{2} + 6 B_{1} + B_{0}) / 6
B_{6} = (21 B_{5}  35 B_{4} + 35 B_{3}  21 B_{2}  7 B_{1}  B_{0}) / 7
(etc.)
The B_{3} and following lines all fit a pattern: the coefficients of the B_{n} come from Pascal's triangle, and the signs alternate, except for the sign of B_{1} which should be inverted (to make it have the same sign as B_{2} and B_{0}).
The temperature (in degrees Celsius) of the Universe at the time the forces began to be distinct, and the highest temperature that can be explained by theory (unless a theory of quantum gravity is developed, which would also explain the universe before the Planck time). The "(85)" is the error range.
(decillion)
One decillion in the "short scale" (billion=10^{9}) system. This is the first numbername that was extrapolated by others based on the names established by Chuquet; see this discussion.
According to Clifford Pickover [190], 10^{33} is the largest power of ten that can be expressed as the product of two numbers that each have no zero digits: 2^{33} × 5^{33} = 8589934592 × 116415321826934814453125 = 10^{33}.
Walt Whitman used decillions, as well as several other large number names in his poetry (see 10^{27} for more). Other numbers mentioned in songs and literature include 525600, 8675309, 10^{10}, and 4.28...×10^{369693099}
See also 10^{63}, 10^{303}, 10^{3003} and 10^{3000003}.
(mass of the Sun)
Approximate mass of the Sun, in grams. See also 5.4×10^{27}.
74994632695013731827926060475067499 ≈ 7.49946...×10^{34}
This is 7499^{9}, and its digits both begin and end with "7499". (from Jim Wilder)
316912650057057350374175801344000001 ≈ 3.169×10^{35}
One of the factors of 10^{10100}+1 (see that entry for details).
419994999149149944149149944191494441 = 648070211589107021^{2}
This number, discovered by Jacobson and Applegate at Carnegie Mellon University, is believed to be the largest square whose digits are all squares (1, 4 and 9). This number is about 4.199×10^{35}; the discoverers determined there was no other smaller than 10^{42}.
Ratio between the electric force and the gravitational force between a proton and an antiproton: (k_{e}e^{2})/(G×m_{p}^{2}), where k_{e} is the Coulomb constant c^{2}/10^{7}, e is the unit charge, G is the gravitational constant, and m_{p} is the mass of the proton using CODATA 2014 values^{50}.
See also the coupling constant 1.693×10^{38} and the "gravityelectric ratio" 2.26881×10^{39}.
The amount (in US dollars) demanded in a lawsuit filed in 2014 May by one Anton Purisima against Au Bon Pain and several other defendants. This broke the previous record. See also 3.01417×10^{15}.
In April 2007, certain websites began publishing an encryption key used in certain types of DVD and BluRay discs. It is 128 binary bits, normally expressed as a 32digit hexadecimal number, but if converted to decimal it is about 1.32562×10^{37}. Under United States law one cannot possess or communicate this key (either explicitly or indirectly), so it has been called an "illegal number".
101097362223624462291180422369532000000 = 54! / ((9!)^{6}) ≈ 1.0109×10^{38}
(Rubik's cube sticker combinations)
This is the number of combinations of the 3×3×3 Rubik's Cube if you're allowed to take the stickers off and replace them in different places. See also 43252003274489856000 and 519024039293878272000.
103919148791293834318983090438798793469 ≈ 1.039191...×10^{38}
The number of legal positions in Go played on a 9×9 board (a popular size for beginners learning the game), computed by John Tromp (with Gunnar Farnebäck and Michal Koucký) in 2005 [231]; see OEIS sequence A94777.
See also 3.724979...×10^{79} and 2.081681...×10^{170}.
115132219018763992565095597973971522401 ≈ 1.151322...×10^{38}
The largest Armstrong or narcissistic number in base 10. It has 39 digits, and is the sum of the 39th powers of its digits. The reason there are no larger numbers is related to the fact that as the number of digits increases more and more 9's are required to get a sum that has N digits. For example, 10^{70}  1 is a number consisting of 70 9's in a row, and the sum of the 70th powers of its digits is 70 × 9^{70} ≈ 4.386051 × 10^{68}, which is only 69 digits long. So there is no way any 70digit number can be equal to the sum of the 70th powers of its digits. The reason we see the last number occur at 39 digits is because, when you get close to the limit, the number of big digits like 8's and 9's has to increase to make sure the sum will be big enough, but this means that there are a lot fewer combinations of digits to choose from. There are 10^{N} Ndigit numbers, but much less than 10^{N} when you start requiring it to have lots of 8's and 9's. The sums of N^{th} powers are fairly evenly distributed, so the overall probability of getting a match decreases. See also 4679307774 and 153.
This is (Mp/n)^{2} where Mp is the Planck mass and n is the mass of the neutron, using CODATA 2014 values^{50}. Its reciprocal is one common version of a gravitational coupling constant, in this case using the neutron as the unit mass. See 3.377×10^{38}.
See also 1.693×10^{38}.
This is (Mp/p)^{2} where Mp is the Planck mass and p is the mass of the proton, using CODATA 2014 values^{50}. Its reciprocal is one common version of a gravitational coupling constant, in this case using the proton as the unit mass.
See also 1.23563×10^{36}, 1.688×10^{38}.
170141183460469231731687303715884105727 = 2^{127}1 ≈ 1.7014...×10^{38}
(a double Mersenne number)
(Lucas' prime record)
C_{4} in Catalan's sequence
This is a Mersenne prime, and the value of C_{4} in Catalan's sequence given by C_{0} = 2; C_{n+1} = 2^{Cn}  1. The sequence starts: 2, 3, 7, 127, and then jumps to this value. All five are prime, and Catalan was conjecturing that all C_{n} are prime.
The primeness of this number was verified by Édouard Lucas in 1876. It was the largest prime ever found by hand calculations, and for 75 years was the largest known prime. This record was beaten in 1951 by Ferrier, who used a mechanical desk calculator; subsequent records have been set by electronic computer. ^{34}
It is not known whether C_{5}=2^{C4}1 is prime.
A Mersenne prime of the form 2^{M}  1 where M is also a Mersenne prime is called a "double Mersenne prime". This is the largest known double Mersenne prime. The next four candidates are 2^{2131}1, 2^{2171}1, 2^{2191}1 and 2^{2311}1, and are all known to be composite.
170141183460469231731687303715884105864 = 2^{127}+136
If you add the numbers in the Catalan sequence 3, 7, 127, 170141183460469231731687303715884105727, you get 3, 10, 137, 2^{127}+136. The last of these is sort of vaguely close to the gravity to electric force ratio (if a ratio of 13 can be thought of as close; see also 10^{40}), and the one before it (137) is famously close to the fine structure constant. Because of these two semicoincidences, some physics theorists have explored the possibility of a Combinatorial hierarchy that can be used to generate most or all of the physical constants and properties of fundamental particles. (See [143], [158], [160] and [161]).
This is twice the "coupling constant" (Mp/n)^{2}=1.688×10^{38}, and a dimensionless combination of physical constants considered significant by reader Mark Thomas. Given:
Mp is the Planck mass
n is the mass of a neutron, approx. 1.67×10^{27} kg,
h is the Planck constant, approx. 6.62×10^{34} m^{2} kg s^{1}
c is the speed of light, 299792458 m s^{1}
G is the gravitational constant, approx. 6.67×10^{11} m^{3} kg^{1} s^{2}
then the expression h c / (π G n^{2}) gives a value close to 70^{2} times the square of the Ramanujan constant e^{π√163}). Thomas connects this to the Leech lattice, the Dirac large numbers hypothesis, and a bunch of other things. See also 137.035 and 3.149544×10^{79}.
3.4028236692093×10^{38} ≈ 2^{128}
This is (approximately) the maximum value that can be represented in the commonlyused IEEE 754 singleprecision (1+8+23 bit) floatingpoint format.
See also 1.797693134862×10^{308}, 1.1897314953572318×10^{4932}, 4.26448742×10^{2525222} and 1.4403971939817846×10^{323228010}.
443426488243037769948249630619149892803 = 27^{27} ≈ 4.434×10^{38}
Just another gratuitous 27 reference. See also 19683.
(gravityelectric force strength ratio)
Ratio between the strength of the gravitational and electric attraction of a proton and an electron. Using CODATA 2014 values^{50} and given
e is the fundamental charge, approx. 1.602×10^{19} C
ε_{0} is the "electric constant", or "vacuum permittivity" constant, 8.854187817...×10^{12} Fm^{1},
G is the Newtonian constant of gravitation, approx. 6.67×10^{11} m^{3}kg^{1}s^{2},
m_{p} is the mass of the proton, approx. 1.67×10^{27} kg,
m_{e} is the [mass of the electron, approx. 9.11×10^{31} kg
the ratio between the strength of the gravitation and electric attraction between the two particles is given by e^{2}/(4πε_{0}Gm_{p}m_{e}). The numbers in parentheses give error ranges.
This number was considered significant by Paul Dirac, see 10^{40}. Dirac put forth this hypothesis at a time when the proton was still considered fundamental (long before the quark model); note that for other similar pairs of particles (e.g. a proton and a muon, or a positron and an electron) you get different ratios. See also 1.23563×10^{36}, 3.377×10^{38} and 1.786×10^{41}.
(Dirac universe constant)
Physicist Paul Dirac's estimate^{70} of the ratio of the universe's size to that of a proton. Using presentday values (see 1.73×10^{15}, 299792458 and 13.72×10^{9}) and a naive assumption that the universe grows at the speed of light, the number would be about 7×10^{39}. Dirac noted that it was "close" to the ratio between the strength of gravity and electical attraction between a proton and an electron (about 2.2687×10^{39}). He hypothesised that it is more than just coincidence, and proposed that the strength of gravity diminishes with time such that the two numbers remain the same. That would mean that when gravity and electricity were of equal strength the universe was about the size of a proton. The choice of the proton's radius is a bit arbitrary; compare to 8.03×10^{60}, and see also 3.377×10^{38}, and 3.1495...×10^{79}.
I have also seen the use of the term "zillion" to refer to this quantity, such as in this quote by Robert Dalling in "The story of us humans ...":
"The electric force is very strong — it is a zillion (10^{45}) times as strong as the gravitational force."
In this context, the time it takes light to travel the distance across a proton (roughly 5.77×10^{24}) is sometimes called a "jiffy".
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Quick index: if you're looking for a specific number, start with whichever of these is closest: 0.065988... 1 1.618033... 3.141592... 4 12 16 21 24 29 39 46 52 64 68 89 107 137.03599... 158 231 256 365 616 714 1024 1729 4181 10080 45360 262144 1969920 73939133 4294967297 5×10^{11} 10^{18} 5.4×10^{27} 10^{40} 5.21...×10^{78} 1.29...×10^{865} 10^{40000} 10^{9152051} 10^{1036} 10^{1010100} — — footnotes Also, check out my large numbers and integer sequences pages.
s.27