what are the prime factors of (105! + 105) ?
where [n!] represents the factorial of the number
$\displaystyle 105! + 105 = 105(104!+1).$
The prime factors of 105 are 3, 5 and 7. As for 104! + 1, it obviously has no prime factors less than 105. According to Sloane, 104! + 1 is not prime, but I have no idea how to determine its factors.
yes i think Soroban is right .
one of the condition that
1. if (a+b) is divided by p then both a and b are multiples of p
2. if (a+b) is divided by p then both a and b are not multiples of p with some more conditions.
for us 1. condition is enough (104!+1) is not divisible by any prime no. till 103
so the only prime factors are 3,7,5 and 104!+1(i am not sure it is prime or not)
Obviously 2 is a prime number in 104!
Won't it be all the prime numbers between 2 and 103? That's possible to do by hand/computer since the list is exhaustive. The product will just be factorised into the same primes/composites again anyway. You could always write some code to recursively calculate all the primes from 2 to 103.
For some bizarre reason I have Fermat's Little Theorem in the back of my head, but my sanity tells me it's not applicable
EDIT: Sorry, I missed the + 1. I'll leave my comments though.
Is there not a theorem by Euclid that states if the product P of finitely many primes plus 1 is not divisible by any prime p in P? Because there will always be a remainder of 1? 104! can be uniquely factorised into finitely many primes like any other number.
Does this help at all?