Sumwise

Factorial of 268

268! = 91726753471968919997811124545451392369123625640642474647572884903647914133430921555354754625480907066969279273264644159857903114627445890834172399995552152106136022107114089263307398546414744668776622509630658979015735325563048887033539620053492927544828514171301673039453176977025806634302116138284662477639208410826443270519874563029892499229380412690715882633111870946335287516085256270438787124296643700216813081573428885726750086476474001969750599649998227548563046400000000000000000000000000000000000000000000000000000000000000000

536 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

268! = 268 × 267 × … × 2 × 1 = 9172675347196891999781112454545139236912…

Facts about 268!

Digits536
Trailing zeros65 (one for every factor 5 × 2)
Stirling estimate≈ Infinity
268! ÷ 267!268

Nearby factorials

266!128188766102030465646222… (532 digits)
267!342264005492421343275414… (534 digits)
269!246744966839596394794111… (539 digits)
270!666211410466910265944102… (541 digits)

What are factorials used for?

Counting arrangements: 268! is the number of different orders in which 268 distinct items can be lined up. Factorials also appear in probability, combinations and series.

Counting with 268!

There are 9.1726 × 10^535 different ways to arrange 268 different books on a shelf: 268 choices for the first place, 267 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 268 runners, there are 268 × 267 × 266 = 19,033,896 possible podiums. If the order does not matter, divide by 3! = 6 to get 3,172,316 ways to choose a group of three. Choosing a pair from 268 people can be done in 35,778 ways.

In probability, if 268 names are drawn out of a hat one at a time, the chance of drawing them in one particular order is 1 in 9.1726 × 10^535. That is why shuffled decks and passwords with many symbols are so hard to guess.

How big is 268!?

268! has 536 digits, so it is larger than the estimated number of atoms in the observable universe (about 10^80). The digits add up to 2133, 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 268!

Every prime up to 268 divides 268!. Legendre's formula counts how often: add ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … for each prime p. For 268! that gives 2^265 × 3^131 × 5^65 × 7^43 × 11^26 × 13^21 × 17^15 × 19^14 × 23^11 × 29^9 × … (56 primes in all). The 65 trailing zeros come from pairing the 65 factors of 5 with factors of 2.

Shuffles where nothing stays put

A related count is the number of derangements of 268 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 268! ÷ e, so about 36.8% of all 9.1726 × 10^535 orders are derangements. From about four items on, that share stays close to 1 ÷ e, roughly 37%, however many items there are.

Computing 268! yourself

In Excel or Google Sheets, type =FACT(268). 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(268) returns the exact integer with all 536 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 268 factorial?

268! is the product of every whole number from 1 to 268.

How many digits does 268! have?

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

How many zeros are at the end of 268!?

65. Count the factors of 5 in the numbers 1 to 268: there are always more factors of 2, so every 5 makes one trailing zero.

Is 268! divisible by 269?

No. 269 is prime and bigger than every factor of 268!, so it cannot divide it (Wilson's theorem says 268! leaves a remainder of 268 when divided by 269).

What is the last non-zero digit of 268!?

4. Strip the 65 trailing zeros and look at the digit that is left at the end.

What is 269! in terms of 268!?

269! = 269 × 268!. Each factorial is the previous one multiplied by the next whole number.