1. number theory problem...

thanks a lot!

2. Originally Posted by Vedicmaths
thanks a lot!
n = 618240007109027021

Select a bound for the factorial.
bound = $\ln(n)$ is an acceptable limit.

In this case about 43 .
The exponent will be 43! (factorial of 43)

Choose some number to raise to that exponential value. Any number (a) from 2 to n-1 is useable.
You may want to check that $\gcd(a,n)=1$.
2 is ok.

$r = ( 2^{43!}\mod n )$

$f = \gcd(r-1,n)$

If
$1
then you have a factor of n
If not then raise the bound limit.

In this case one of the factors is a 9 digit number.
The first 3 digits and the last 3 digits are
250___201

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