Sumwise

Factorial of 263

263! = 68884027232548817179216648527172473578241428906973375771263803047752982112422438306381984658745965404339944344730188908707054928103119659421645829757344140046675780471670685132302677559040654341644772727373251331486845971768947393535842517102278292668929181160302116536912370726148164141169352378474870162705542354318295299430132615720956914808987246162670870122646784296076195117776919971970500492650769854343695661700914697101220994791899840532636536710103040000000000000000000000000000000000000000000000000000000000000000

524 digits

Sumwise · Maintained by Theo Abbott (pen name) · automated checks · Data: exact arithmetic; unit definitions from NIST SP 811 · Updated · How the checks work

How it is calculated

263! = 263 × 262 × … × 2 × 1 = 6888402723254881717921664852717247357824…

Facts about 263!

Digits524
Trailing zeros64 (one for every factor 5 × 2)
Stirling estimate≈ Infinity
263! ÷ 262!263

Nearby factorials

261!999681119678240170365666… (519 digits)
262!261916453355698924635804… (522 digits)
264!181853831893928877353131… (527 digits)
265!481912654518911524985799… (529 digits)

What are factorials used for?

Counting arrangements: 263! is the number of different orders in which 263 distinct items can be lined up. Factorials also appear in probability, combinations and series.

Counting with 263!

There are 6.8884 × 10^523 different ways to arrange 263 different books on a shelf: 263 choices for the first place, 262 for the second, and so on down to one choice for the last. If you only fill the first three places, as with gold, silver and bronze among 263 runners, there are 263 × 262 × 261 = 17,984,466 possible podiums. If the order does not matter, divide by 3! = 6 to get 2,997,411 ways to choose a group of three. Choosing a pair from 263 people can be done in 34,453 ways.

In probability, if 263 names are drawn out of a hat one at a time, the chance of drawing them in one particular order is 1 in 6.8884 × 10^523. That is why shuffled decks and passwords with many symbols are so hard to guess.

How big is 263!?

263! has 524 digits, so it is larger than the estimated number of atoms in the observable universe (about 10^80). The digits add up to 2061, a multiple of 9, as they are for every factorial from 6! on. Stirling's formula, √(2πn) × (n/e)^n, gives a close estimate without multiplying everything out, and gets more accurate as n grows.

Prime factors of 263!

Every prime up to 263 divides 263!. Legendre's formula counts how often: add ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … for each prime p. For 263! that gives 2^259 × 3^129 × 5^64 × 7^42 × 11^25 × 13^21 × 17^15 × 19^13 × 23^11 × 29^9 × … (56 primes in all). The 64 trailing zeros come from pairing the 64 factors of 5 with factors of 2.

Shuffles where nothing stays put

A related count is the number of derangements of 263 items: orders in which no item ends up in its original place, as in a Secret Santa where nobody draws their own name. The count is the whole number nearest to 263! ÷ e, so about 36.8% of all 6.8884 × 10^523 orders are derangements. From about four items on, that share stays close to 1 ÷ e, roughly 37%, however many items there are.

Computing 263! yourself

In Excel or Google Sheets, type =FACT(263). Spreadsheets keep only 15 significant digits, so from 18! on the result is rounded, and FACT stops at 170!, the largest factorial that fits in a standard floating-point number. In Python, math.factorial(263) returns the exact integer with all 524 digits, because Python integers have no size limit. The gamma function extends factorials to fractions: Γ(n + 1) = n!, so (1/2)! = √π ÷ 2 ≈ 0.8862.

Why is 0! equal to 1?

There is exactly one way to arrange nothing, and the rule n! = n × (n − 1)! only works for n = 1 if 0! = 1. Defining it this way keeps formulas for combinations and series correct at the edges.

Calculate another factorial

How answers are calculated and checked: methodology.

Questions

What is 263 factorial?

263! is the product of every whole number from 1 to 263.

How many digits does 263! have?

524. The number of digits of n! is the whole part of log10(n!) plus one.

How many zeros are at the end of 263!?

64. Count the factors of 5 in the numbers 1 to 263: there are always more factors of 2, so every 5 makes one trailing zero.

Is 263! divisible by 264?

Yes. 264 is not prime, so its factors already appear among the numbers 1 to 263.

What is the last non-zero digit of 263!?

4. Strip the 64 trailing zeros and look at the digit that is left at the end.

What is 264! in terms of 263!?

264! = 264 × 263!. Each factorial is the previous one multiplied by the next whole number.