Evaluating 993!

Computer Science Level 3

How many digits are there in 993!?

The answer is 2547.

8 solutions

Shaun Loong
Apr 22, 2014

To solve this question without computer programming, we take the logarithm of 993 factorial. Since the number of digits of $x$ can be calculated using the formula $d=\lceil y \rceil+1$ , where $d$ is the number of digits of $x$ and $y=\log_{10}x$ . $\log_{10}993!=\sum_{n=1}^{993}\log_{10}n=2546.613...$

$\therefore$ number of digits in $993!$ is $2546+1=\boxed{2547}$ .

Technically, aren't you using computing to find what $\sum \log_{10} n$ is?

You can approximate it by using integration, to evaluate it as $\int_1^{994} \log_{10} x \, dx > \sum \log_{10} n > \int_1^{993} \log_{10} x \, dx$ . This gives us $2548.15 > \sum > 2545.15$ . It could be improved slightly by accounting for the first few terms (which give the greatest error). In any case, this gets us close to the answer without much work.

Calvin Lin Staff - 7 years, 1 month ago

It can also be done purely analytically without any use of computation by using Stirling's approximation (which gives both a lower and matching upper bound) $\log_{10}(n!)\asymp \frac{1}{2}\log_{10}(2 \pi n) + n \log_{10} (n/e)$

Abhishek Sinha - 5 years, 2 months ago

hi.shaun well I knew this formula but i still did not understand how you calculated the log of 993! please explain in detail

Max B - 7 years, 1 month ago

it follows as the follwing: log 4! = log (4 3 3 2 1) = log 4 + log 3 + log 2 + log 1
{as, log (m*n)= log m + log n } there fore ... the problem follows the ceiling function.. to get the result of the problem

NEELESH SRIVASTAVA - 7 years, 1 month ago

It should actually be $d=\lfloor y\rfloor +1$ . Instead of remembering the formula incorrectly, try to figure out how the formula works. The basic idea is that $y=f(x)=\log_{10}x$ is strictly increasing on $(0,\infty)$ and,

$\forall~x\in\Bbb{Z^+}~,~y=f(x)=m\in\Bbb{Z_{\geq 0}}\iff x=\min(A_{m+1})$

where $A_i$ denotes the set of all $i$ -digit positive integers $\forall~i\in\Bbb{Z^+}$ .

Prasun Biswas - 5 years, 4 months ago

Scott Handelman
Apr 27, 2014

Is it cheating to use built-in modules? Python solution:

from math import factorial
print len(str(factorial(993)))

Cool! I didn't think of using len and I used log.....

展豪張 - 5 years, 2 months ago

Daniel Gabrić
Apr 22, 2014

Easy stuff

public class factorialDigits {
    public static void main(String[] args) {
        BigInteger f = factorial(993);
        System.out.println(f.toString().length());
    }
    public static BigInteger factorial(int n) {
        BigInteger prod = BigInteger.ONE;
        for (int i = 2; i <= n; i++) {
            prod = prod.multiply(BigInteger.valueOf(i));
        }
        return prod;
     }
}

Jesse Nieminen
Mar 5, 2016

Using Java:

//Defined for non-negative integers
public static BigInteger factorial(int number) {
    BigInteger factorial = BigInteger.ONE; //Equals 1

    //Loop through all integers from 2 to number (we can skip 1 since multiplying by 1 does nothing)
    for (int i = 2; i <= number; i++) {

        BigInteger current = new BigInteger(String.valueOf(i)); //Equals i

        factorial = factorial.multiply(current); //Multiply factorial by the current number

    }

    return factorial; //Return factorial
}

public static int length(BigInteger number) {

    return number.toString().length(); //Get the number as a string and get its length

}

public static void main(String[] args) {

    System.out.println(length(factorial(993))); //Compute the factorial and print it

}

Which prints out $\boxed{2547}$

Anubhav Balodhi
Jun 2, 2015

I wrote a code in Python: (http://ideone.com/VebyRT)

1
2
3

from math import factorial
x=factorial(993)
print(len(str(x)))

Kanav Gupta
Apr 28, 2014

To calculate the value , I just simply used this code in Python-

def f(n):
  if n==0:
    return 1
  return n*f(n-1)

x=str(f(993))

print len(x)

This gave me solution 2547

I did exactly the same ^_^ :D

Krishna Ar - 6 years, 7 months ago

Masbahul Islam
Apr 26, 2014

Python code:

def masba(n):

if n==1:

  return 1

else:

  return n*masba(n-1)

print len(str(masba(993)))

Aditya Mishra
Apr 25, 2014

def fact(i):

if(i==1):

    return 1

else:

    return i*fact(i-1)

x=fact(993)

i=0

while(x>0):

x//=10

i+=1

print(i)

Evaluating 993!

The answer is 2547.

8 solutions

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