What Is the Biggest Number Anyone Has Used?

Play with the digit before reading on.

• a.With three 9s, put the five in order from smallest to biggest. Where does $(9^9)^9$ come?
• b.Now try three 2s. Which wins? Why does the tower do so badly?
• c.Try three 3s. Who wins now, and by how much?
• d.From which digit on does the tower always win?
• e.$9^{9^{\scriptstyle 9}}$ and $(9^9)^9$ are made of the same three 9s. How many digits does each have?

So two things decide which number is biggest: the power matters far more than the number being raised, and a tower puts a power in the power. But the digits must be big enough to grow: with 2s and 3s the tower is still too short to take off.

Nobody has ever written out $9^{9^{\scriptstyle 9}}$ by hand. So how do we know it has 369,693,100 digits, and how can we compare it with another number we can’t write either?

Counting digits with logarithms

Every number of 1 or more sits between two powers of 10. A number from 1,000 to 9,999 is at least 10³ and less than 10⁴, and has 4 digits. So a number’s number of digits is one more than the whole-number part of the power of 10 it equals. That power is its logarithm, $\log N$ (to base 10): $\log 1{,}000 = 3$, $\log 5{,}000 \approx 3.7$, and 5,000 has 3 + 1 = 4 digits.

The logarithm of a power is easy, because raising to a power becomes multiplying: $\log (a^b) = b \log a$. With $\log 9 \approx 0.9542$:

• $\log (9^{99}) = 99 \times 0.9542 = 94.47$, so 9⁹⁹ has 95 digits.
• $\log (99^9) = 9 \times \log 99 = 9 \times 1.9956 = 17.96$, so 99⁹ has 18 digits.
• $\log \left(9^{9^{\scriptstyle 9}}\right) = 387{,}420{,}489 \times 0.95424250944 = 369{,}693{,}099.63$, so the tower has 369,693,100 digits. Here $\log 9$ needs 11 decimal places, because it is multiplied by almost 40 crore: with 0.954243 the answer would come out about 190 too big.

The part after the point tells you how the number begins. $10^{0.6316} \approx 4.28$, so the tower is about $4.28 \times 10^{369{,}693{,}099}$, written in standard form (scientific notation): a number from 1 to 10 times a power of 10. That one line says what 123,000 pages of digits would, at 3,000 digits a page.

When even the digits are too many: the log of the log

For a tower, even the logarithm is too big to write easily, so we take a logarithm again. The tower of four 3s, $3^{3^{\scriptstyle 3^{\scriptstyle 3}}}$, is 3 raised to $3^{27} = 7{,}625{,}597{,}484{,}987$. Its logarithm is $7{,}625{,}597{,}484{,}987 \times 0.4771 \approx 3.64 \times 10^{12}$, so it has about 3.6 trillion digits (3,638,334,640,025 exactly). Printed at 3,000 digits a page on paper 0.1 mm thick, that is a stack about 120 km high. Add one more 3 at the bottom and the number of digits itself has 3.6 trillion digits.

That is why mathematicians who work with such numbers describe them by how many times you must take the logarithm to bring them down to an ordinary size. A googol, $10^{100}$, needs it twice (100, then 2). $9^{9^{\scriptstyle 9}}$ needs it twice to reach about 8.6. Each extra level of a tower needs one more.

Numbers this big were first written down for a reason that sounds like a joke. In the 3rd century BCE, the Greek mathematician Archimedes of Syracuse wrote a short book, The Sand Reckoner, to the king of Syracuse, Gelon. It begins (in Thomas Heath’s translation of 1897): “There are some, king Gelon, who think that the number of the sand is infinite in multitude.” Others, he says, think it is finite but bigger than any number that has a name. He set out to prove both wrong by counting the grains of sand it would take to fill the whole universe.

The Greeks’ biggest number with its own name was the myriad, 10,000. Archimedes built on it. A myriad myriads, $10^8$, ends the first order of numbers. Then $10^8$ becomes the unit of the second order, which counts up to $10^8 \times 10^8 = 10^{16}$; and so on, order after order, a myriad myriad orders of them, up to $(10^8)^{10^{\scriptstyle 8}} = 10^{800{,}000{,}000}$. All that is only the first period. He went on for a myriad myriad periods, to $10^{8 \times 10^{\scriptstyle 16}}$: a 1 followed by 80,000,000,000,000,000 zeros. Along the way he proved the rule you use for indices, that multiplying powers of 10 adds their powers.

How big is a googol?

A googol is more than anything you can count in the universe. The observable universe holds about $10^{80}$ atoms; filled solid with sand, it would hold about $10^{95}$ grains. So a googol is $10^{100} \div 10^{80} = 10^{20}$ times the number of atoms, and a googolplex can’t even be written down: it has a googol + 1 digits, and there are not enough atoms to write them on, even one digit per atom.

Towers get tall quickly, and soon there is no room to draw them. In 1976 the computer scientist Donald Knuth gave them a short notation, up-arrows. One arrow is a power, $3 \uparrow 3 = 3^3 = 27$. Two arrows are a tower: $3 \uparrow\uparrow 4$ is a tower of four 3s.

Each new operation repeats the one before. Multiplying is repeated adding, 3 × 4 = 3 + 3 + 3 + 3; a power is repeated multiplying, $3^4 = 3 \times 3 \times 3 \times 3$; a tower is repeated powering. Three arrows repeat towers: $3 \uparrow\uparrow\uparrow 3$ is $3 \uparrow\uparrow (3 \uparrow\uparrow 3)$, a tower of 3s that is 7,625,597,484,987 levels high. A tower of only five 3s already has a number of digits with 3.6 trillion digits of its own.

Written	What it means	How big
$3 \uparrow 3$	$3^3$	27
$3 \uparrow\uparrow 3$	$3^{3^{\scriptstyle 3}}$	7.6 trillion
$3 \uparrow\uparrow 4$	a tower of four 3s	3.6 trillion digits
$3 \uparrow\uparrow\uparrow 3$	a tower of 3s, 7.6 trillion high	past counting digits
$3 \uparrow\uparrow\uparrow\uparrow 3$	$3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow\uparrow 3)$	the first step to Graham’s number

The number in a proof

In 1971 Ronald Graham and Bruce Rothschild worked on a question about coloring. Take a cube in n dimensions (a square has 4 corners, a cube 8, a 4-dimensional cube 16, an n-dimensional one $2^n$). Join every pair of corners with a line, and color each line red or blue, any way you like. Is there always a set of four corners, lying flat in one plane, whose six lines are all one color? For small n you can color to avoid it; for big enough n you can’t. They proved that the smallest such n exists and gave a limit it can’t be above.

Graham explained a simpler, bigger limit to the writer Martin Gardner, who in 1977 called it the largest number ever used in a serious mathematical proof. Start with $g_1 = 3 \uparrow\uparrow\uparrow\uparrow 3$. Then g₂ is 3, then g₁ arrows, then 3. Then g₃ has g₂ arrows, and so on, 64 times: Graham’s number is $g_{64}$. The answer to the coloring question is at least 13 (found by 2008), and since 2014 its limit has been brought down far below Graham’s number. So Graham’s number was a ceiling, and a very loose one.

Graham’s number is far too big for its digits ever to be written, or even counted. And yet its last digits are known exactly: it ends …⁠2464195387. How can anyone know how a number ends without being able to write it? Watch the end of a tower of 3s as it grows.

Each 3 added to the tower settles one more digit at the end. A tower of 13 threes ends …⁠262464195387, and so does every taller tower of 3s, Graham’s number among them (it is one: its last step is 3 raised to an enormous tower of 3s).

Why the last digit is always 7

Look at the last digit of the powers of 3: 3, 9, 27, 81, 243, 729, 2,187, 6,561… The last digits go 3, 9, 7, 1, 3, 9, 7, 1: every fourth power the pattern starts again, because the last digit of a product depends only on the last digits multiplied, and $3^4 = 81$ ends in 1, so four more 3s multiply the last digit by something ending in 1, which leaves it as it was. So the last digit of $3^x$ depends only on the remainder when x is divided by 4: remainder 1 gives 3, 2 gives 9, 3 gives 7, 0 gives 1.

In a tower of 3s (two or more), x is itself a power of 3 with an odd power. What does an odd power of 3 leave when divided by 4? 3 = 4 − 1, so its powers alternate: 3 leaves 3, 9 leaves 1, 27 leaves 3, 81 leaves 1. An odd power always leaves 3. So the tower’s last digit is the third in the cycle: 7, every time, however tall.

The next digits follow from the same idea with bigger cycles: the last two digits of the powers of 3 repeat every 20 powers, the last three every 100, the last four every 500. Working down the tower one level at a time, each cycle a size the level below can be reduced to, a computer finds the last 500 digits of Graham’s number in moments. This is modular arithmetic, the arithmetic of remainders, and the rule that says how long each cycle can be is Euler’s theorem (Leonhard Euler, 1763).

No. Whatever number anyone names, add 1. So “the biggest number” never has an answer, and “the biggest number anyone has used” keeps changing. Gardner’s record didn’t last: numbers far larger than Graham’s number have since appeared in serious proofs, among them TREE(3), which comes from a theorem about trees (the branching drawings, not the plants).

There are even numbers that are perfectly well defined and yet no computer program can work them all out. In 1962 Tibor Radó took the simplest kind of computer, a Turing machine, with n states (a state is like a numbered line of a program, with a rule for each symbol it reads), and asked: of all such machines that eventually stop, how long does the longest-running one run? These are the busy beaver numbers. For 5 states the answer is 47,176,870 steps, proved only in 2024 by an online collaboration, the Busy Beaver Challenge, many of them amateurs, with the proof checked by computer. For 6 states it is more than $2 \uparrow\uparrow\uparrow 5$ (as of 2026). Radó proved that the busy beaver numbers grow faster than anything any program can calculate, so no program can find them all.

And however big a number gets, it is still nowhere near infinity: $9^{9^{\scriptstyle 9}}$ and Graham’s number are just as far from it as 1 is. Infinity is not a very big number but a different kind of thing, with surprises of its own.

• Locks for computers. The AES cipher allows keys of 256 bits: $2^{256} \approx 1.16 \times 10^{77}$ possible keys. A supercomputer of 2019 checking $10^{14}$ keys a second would need about $3.67 \times 10^{55}$ years to try them all. A code is safe by being a big number.
• Prime numbers. The largest prime known (as of October 2026) is $2^{136{,}279{,}841} - 1$, with 41,024,320 digits, found in October 2024 by the Great Internet Mersenne Prime Search on thousands of graphics processors. Its size is known from its logarithm, as in this topic.
• Games. Claude Shannon estimated in 1950 that there are at least $10^{120}$ possible games of chess (about $10^3$ choices for each pair of moves, over 40 pairs), more than the atoms in the universe, so no computer can play chess by checking every game.
• Primes and Skewes’s number. For small numbers, a famous estimate of how many primes there are below a number always comes out too high. In 1914 J. E. Littlewood proved it doesn’t stay that way, and in 1933 Stanley Skewes showed (assuming a famous unproved conjecture, the Riemann hypothesis) that the first exception comes before $10^{10^{\scriptstyle 10^{\scriptstyle 34}}}$. It is now known to come before about $1.4 \times 10^{316}$, far smaller.

• Why Is a Magnitude 8 Earthquake Not Just a Bit Bigger Than a 7? Logarithms and log scales, which this topic used to count digits.
• How Did a Book of Numbers Make Multiplying Easy? Napier’s logarithms and the slide rule.
• What Don’t Mathematicians Know About Prime Numbers? Where Skewes’s number came from, and the questions still open.
• Are Some Infinities Bigger Than Others? Past every number, to infinity, and the surprise that one infinity can be bigger than another.
• Grades 6–8: How Much Rice Is on the Chessboard?, How Long Is a Crore Seconds? (Nepal’s names for big numbers, and the prince who counted), Is There a Biggest Prime Number? and Is There Room at a Full Hotel with Endless Rooms?.
• Computing, Going Further: How Many Moves to Move 64 Disks? and How Does a Computer Play Chess?; and What Is the Biggest Number a Computer Can Count To? (Coming soon.)
• Grade 10, Indices.