Shoutout To Number Theory Enthusiasts!

Computer Science Level 3

Find the 40th digit from the left of the expansion of 123 ! 123! .

Details and Assumptions:

  • ! ! denotes the factorial operator. (For example, 7 ! = 7 × 6 × 5 × . . . × 2 × 1 7! = 7 \times 6 \times 5 \times ... \times 2 \times 1 )
  • As an example, the 9th digit from the left of 12342433 2 432 is 2.
4 7 6 2 1 3 9 5

Arkajyoti Banerjee by Arkajyoti Banerjee , 20, India

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2 solutions

Fletcher Mattox
May 1, 2021 
1
2
 
 from math import factorial
 print(str(factorial(123))[39])


1
 
5

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A purely mathematical solution would've worked: 

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import math

#Defining the factorial function using recursion
def fact(n):
    if n == 0:
        return 1
    else:
        return n * fact(n-1)

n=fact(123)
j=0
k=40
temp=n
while(temp >= 1):
    j += 1
    temp /= 10
temp=n
temp = math.floor(temp/10**(j-k))
print(temp % 10)

But it doesn't show proper results for any digit past the 1 5 t h 15^{th} digit of a number consisting of more than or equal to 16 16 digits.

So, we convert the number to a string datatype and print the character from the required position: 

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#Defining the factorial function using recursion
def fact(n):
    if n == 0:
        return 1
    else:
        return n * fact(n-1)

n=fact(123)
k=40
#Converting the number into a string
n=str(n)

#n[k] refers to the character in the kth index of the array of characters of the string n filled up in the same order as in the string
#array indices start from 0,1,2,3,... which represent positions 1,2,3,4,... respectively
#So the character at position k must be at index (k-1), so the character at the (k-1)th index of the character array n is the
#digit at position k

print(n[k-1])

Alternative C++ equivalent: (Although this too can't process very large values and is evidently not gonna work here.) 

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#include<iostream>
#include<cmath>
using namespace std;

main()
{
    int n,k,j,temp,rem;
    cout<<"Enter a number : ";
    cin>>n;
    input:
    cout<<"Enter the position of the digit : ";
    cin>>k;
    temp=n;
    j=0;
    while(temp >= 1)
    {
        j++;
        temp /= 10;
    }
    if(k > j)
    {
        cout<<"\nThe entered number has only "<<j<<" digits. Try again.\n";
        goto input;
    }
    temp=n;
    temp /= pow(10,j-k);
    temp=floor(temp);
    rem=temp%10;
    cout<<"The digit at position "<<k<<" of "<<n<<" is "<<rem;
}

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