It's hard to post code here but I think this works.

// most code stolen from
// https://www.geeksforgeeks.org/segmented-sieve-print-primes-in-a-range/

#include <bits/stdc++.h>

#include "BigInt.hpp"  // https://faheel.github.io/BigInt/

using namespace std;
// fillPrime function fills primes from 2 to sqrt of high in chprime(vector) array
void fillPrimes(vector<int>& prime, int high)
{
    bool ck[high + 1];
    memset(ck, true, sizeof(ck));
    ck[1] = false;
    ck[0] = false;
    for (int i = 2; (i * i) <= high; i++) {
        if (ck[i] == true) {
            for (int j = i * i; j <= sqrt(high); j = j + i) {
                ck[j] = false;
            }
        }
    }
    for (int i = 2; i * i <= high; i++) {
        if (ck[i] == true) {
            prime.push_back(i);
        }
    }
}
// in segmented sieve we check for prime from range [low, high]
void segmentedSieve(int low, int high)
{
    if (low<2 and high>=2){
        low = 2;
    }//for handling corner case when low = 1 and we all know 1 is not prime no. 
    bool prime[high - low + 1];
  //here prime[0] indicates whether low is prime or not similarly prime[1] indicates whether low+1 is prime or not
    memset(prime, true, sizeof(prime));
 
    vector<int> chprime;
    fillPrimes(chprime, high);
    //chprimes has primes in range [2,sqrt(n)]
     // we take primes from 2 to sqrt[n] because the multiples of all primes below high are marked by these 
   // primes in range 2 to sqrt[n] for eg: for number 7 its multiples i.e 14 is marked by 2, 21 is marked by 3,
   // 28 is marked by 4, 35 is marked by 5, 42 is marked 6, so 49 will be first marked by 7 so all number before 49 
  // are marked by primes in range [2,sqrt(49)] 
    for (int i : chprime) {
        int lower = (low / i);
        //here lower means the first multiple of prime which is in range [low,high]
        //for eg: 3's first multiple in range [28,40] is 30          
        if (lower <= 1) {
            lower = i + i;
        }
        else if (low % i) {
            lower = (lower * i) + i;
        }
        else {
            lower = (lower * i);
        }
        for (int j = lower; j <= high; j = j + i) {
            prime[j - low] = false;
        }
    }
   

    BigInt numerator, denominator, remainder;

    // You'll need to set your terminal to 60 columns for this 
    // to look right. 
    numerator = 
    "111111111111111111111111111111111111111111111111111111111111" //line 01
    "111111111111111111111111111111111111111111111111111111111111" //line 02
    "111111111111111111111111111111111111111111111111111111111111" //line 03 
    "111111111111111111111111111111111111111111111111111111111111" //line 04
    "111111111111111111111111111111111111111111111111111111111111" //line 05
    "111111111111111111111111111111111111111111111111111111111111" //line 06 
    "111111111111188888888888111111111111111111888811111111111111" //line 07 
    "111111111111118888888888881111111111111188888111111111111111" //line 08 
    "111111111111111888811118888111111111118888811111111111111111" //line 09 
    "111111111111111118888111888811111111188881111111111111117111" //line 10 // originally ends 7111
    "111111111111111111888811188888111118888811111111111111111111" //line 11 
    "111111111111111111188888111888811888881111111111111111111111" //line 12 
    "111111111111111111111888811188888888111111111111111111111111" //line 13 
    "111111111111111111111188881118888811111111111111111111111111" //line 14 
    "111111111111111111111118888811188881111111111111111111111111" //line 15 
    "111111111111111111111111188881118888811111111111111111111111" //line 16 
    "111111111111111111111111188888811188881111111111111111111111" //line 17 
    "111111111111111111111118888888881118888111111111111111111111" //line 18 
    "111111111111111111111888881118888111888881111111111111111111" //line 19 
    "111111111111111111118888111111888881118888111111111111111111" //line 20 
    "111111111111111111888881111111118888111888811111111111111111" //line 21 
    "111111111111111188888111111111111888811118888111111111111111" //line 22 
    "111111111111118888811111111111111188888888888811111111111111" //line 23 
    "111111111111188881111111111111111111888888888881111111111111" //line 24 
    "111111111111111111111111111111111111111111111111111111111111" //line 25 
    "111111111111111111111111111111111111111111111111111111111111" //line 26 
    "111111111111111111111111111111111111111111111111111111111111" //line 27 
    "111111111111111111111111111111111111111111111111111111111111" //line 28 
    "111111111111111111111111111111111111111111111111111111111111" //line 29 
    "111111111111111111111111111111111111111111111111111111111111"; //line 30
	    

    int primecount=0;
    BigInt root;
    root = sqrt(numerator);
    cout << "A prime factor might be as large as sqrt(n) " << endl << root << endl;
    cout << "Looking for factors for " << endl << numerator << endl;
    for (int i = low; i <= high; i++) {
        if (prime[i - low] == true) {
	    primecount++;
	    denominator = i;
	    remainder = numerator%denominator;
	    if( remainder==0 )
            {
            	cout << endl << (i) << " * " <<endl<< numerator/denominator << "\n";
	    }
        }
    }
    cout << "Found " << primecount << " primes from "<< low << " to " << high << endl;
}
int main()
{
    // low should be greater than or equal to 2
    int low = 2;
    // low here is the lower limit
    int high = 1001000;
    // high here is the upper limit
      // in segmented sieve we calculate primes in range [low,high] 
   // here we initially we find primes in range [2,sqrt(high)] to mark all their multiples as not prime
   //then we mark all their(primes) multiples in range [low,high] as false
   // this is a modified sieve of eratosthenes , in standard sieve of  eratosthenes we find prime from 1 to n(given number)
   // in segmented sieve we only find primes in a given interval 
  cout<<"Primes in range "<<low<<" to "<< high<<" are\n";
    segmentedSieve(low, high);
}

Thanks.

To be sure, we need to check all primes up to ~10^900. The number of factors to check is given by the prime counting function π(n), a lower bound of which can be estimated by calculating n/ln(n), where ln(x) is the natural log.

Doing this, you see the number of primes that need to be checked is on the order of 10^895. This is literally impossible in the age of the universe even with several supercomputers doing trillions of multiplies per second. So we are left with 2 options:

1. Elon lied and he has no idea if that number is prime or not. He just found a number that couldn't be easily detected.

2. Elon has access to an extremely advanced quantum computer that can run Shor's algorithm on a number that large.

I love that you did this. I was thinking of doing same, but I haven’t written code in years and don’t have an environment set up.

The square root
333333333333333333333333333333333333333333333333333333333333
333333333333333333333333333333333333333333333333333333333333
333333333333333333333333333333333333333333333333333333333333
333333333333333333333333333333333333333333333333333333333333
333333333333333333333333333333333333333333333333333333333333
333333333333333333333333333333333333333333333333333333333333
333333333333449999999998833333333333333334499883333333333333
333333333333344999999999988333333333333449998833333333333333
333333333333334499883344998833333333344999883333333333333333
333333333333333344998834499883333333449988333333333333342333
333333333333333334499883449998833344999883333333333333333333
333333333333333333449998834499884499988333333333333333333333
333333333333333333334499863033333332574999999997958332925040
833337416258333333333447943003312800044916666666625792459579
083741703412583333332936710853915817588232925040824958453458

The square root squared
111111111111111111111111111111111111111111111111111111111111
111111111111111111111111111111111111111111111111111111111111
111111111111111111111111111111111111111111111111111111111111
111111111111111111111111111111111111111111111111111111111111
111111111111111111111111111111111111111111111111111111111111
111111111111111111111111111111111111111111111111111111111111
111111111111188888888888111111111111111111888811111111111111
111111111111118888888888881111111111111188888111111111111111
111111111111111888811118888111111111118888811111111111111111
111111111111111118888111888811111111188881111111111111117111
111111111111111111888811188888111118888811111111111111111111
111111111111111111188888111888811888881111111111111111111111
111111111111111111111888811188888888111111111111111111111111
111111111111111111111188881118888811111111111111111111111111
111111111111111111111118888811188881111111111111111111111110
872829766805272222263052288947413672766557783219255569213941
447307699280527225195557780774127451227375356265731617644147
719473319952550462768856734722303882556239158668425548064747
746447307497686593445540272222228284367434645476762884218253
573255919484013293323952622207934436929984639266847531918647
934058248134047178290682424792993766675742027197004790912104
070880609576268319006407829811241555751080534940879976181963
493032008011917914713109819286427923400472101972124569141732
386089979846058899306154373119593290630091049006267261249144
384834907967504235972651612315027493087135219480813590851065
858223354310021351183469186819960345709802584665828955609783
783283614995288483833873044084153826152922130050145045452436
393000830600802484169201670819087221448461390917691632988949
889920519658137805513502850195873988200632075141289958806068
116255494495628879568891996083540200828642026571815152157764

Elon: My super computer can beat up your super computer."

Can someone please direct me to the original Elon (graphical) post? I am having a tough time finding it......

I had already done this in Python yesterday. It is a prime number.

I was also scratching my head on this, how can he generate that prime number while embedding an X logo as well as a "17" in a 60x30 matrix of numbers. I tried to generate an 1800 digit prime number, all I got was the first prime number containing 1800 digits that started with a "1", ending with "1953" and all "0"s in between, using below python script (as suggested by AI):

import sympy

prime_number = sympy.nextprime(10**1799)

print(f"A prime number with 1800 digits:\n{prime_number}")

It would require a lot of computing power to get what Elon obtained (he probably used Grok for this).