Sumwise

Factorial of 273

273! = 134064227083159708398531635042001113901224662136819467042366772112801937207102940623738907312224249736986657475050499143940843303496530029781566282541146111878947439020819009234983192090512484977908381387443976431111083278520566617141235040439514791516849520184811074417023743100895861350807004874345871843669521581338311667490027819259360594494213020339666795780355304040737264430344058004812043037525260424591604556712095329603657085776664426644459380018810457846461830092430508032000000000000000000000000000000000000000000000000000000000000000000

549 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

273! = 273 × 272 × … × 2 × 1 = 1340642270831597083985316350420011139012…

Facts about 273!

Digits549
Trailing zeros66 (one for every factor 5 × 2)
Stirling estimate≈ Infinity
273! ÷ 272!273

Nearby factorials

271!180543292236532682070851… (544 digits)
272!491077754883368895232716… (546 digits)
274!367335982207857601011976… (551 digits)
275!101017395107160840278293… (554 digits)

What are factorials used for?

Counting arrangements: 273! is the number of different orders in which 273 distinct items can be lined up. Factorials also appear in probability, combinations and series.

Counting with 273!

There are 1.3406 × 10^548 different ways to arrange 273 different books on a shelf: 273 choices for the first place, 272 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 273 runners, there are 273 × 272 × 271 = 20,123,376 possible podiums. If the order does not matter, divide by 3! = 6 to get 3,353,896 ways to choose a group of three. Choosing a pair from 273 people can be done in 37,128 ways.

In probability, if 273 names are drawn out of a hat one at a time, the chance of drawing them in one particular order is 1 in 1.3406 × 10^548. That is why shuffled decks and passwords with many symbols are so hard to guess.

How big is 273!?

273! has 549 digits, so it is larger than the estimated number of atoms in the observable universe (about 10^80). The digits add up to 1980, 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 273!

Every prime up to 273 divides 273!. Legendre's formula counts how often: add ⌊n/p⌋ + ⌊n/p²⌋ + ⌊n/p³⌋ + … for each prime p. For 273! that gives 2^270 × 3^135 × 5^66 × 7^44 × 11^26 × 13^22 × 17^16 × 19^14 × 23^11 × 29^9 × … (58 primes in all). The 66 trailing zeros come from pairing the 66 factors of 5 with factors of 2.

Shuffles where nothing stays put

A related count is the number of derangements of 273 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 273! ÷ e, so about 36.8% of all 1.3406 × 10^548 orders are derangements. From about four items on, that share stays close to 1 ÷ e, roughly 37%, however many items there are.

Computing 273! yourself

In Excel or Google Sheets, type =FACT(273). 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(273) returns the exact integer with all 549 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 273 factorial?

273! is the product of every whole number from 1 to 273.

How many digits does 273! have?

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

How many zeros are at the end of 273!?

66. Count the factors of 5 in the numbers 1 to 273: there are always more factors of 2, so every 5 makes one trailing zero.

Is 273! divisible by 274?

Yes. 274 is not prime, so its factors already appear among the numbers 1 to 273.

What is the last non-zero digit of 273!?

2. Strip the 66 trailing zeros and look at the digit that is left at the end.

What is 274! in terms of 273!?

274! = 274 × 273!. Each factorial is the previous one multiplied by the next whole number.