Thread: stuck with factors of (105! + 105)

1. stuck with factors of (105! + 105)

what are the prime factors of (105! + 105) ?

where [n!] represents the factorial of the number

2. prime factors are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61, 67,71,73,79,83,89,97,101,103.

i think there is no shortcuts.

3. The prime factors of 105! are the primes less than 105...

Kind regards

$\chi$ $\sigma$

4. i wanted prime factors of 105! + 105 and not just 105!

5. Hello, grgrsanjay!

Originally Posted by grgrsanjay
i wanted prime factors of $105! + 105$ and not just $105!$

$105! + 105 \;=\;105\cdot(104! + 1)$

The prime factors are: . $3,\,5,\,7,\,\text{ and }\,what?$

6. and what?

hmm...

i could not get you still..

7. Originally Posted by grgrsanjay
what are the prime factors of (105! + 105) ?
$105! + 105 = 105(104!+1).$

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.

8. so, there isint any way?

9. 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)

10. 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.

11. Originally Posted by grgrsanjay
and what?

hmm...

i could not get you still..
Surely it's obvious that (104! + 1) is the "what" that was refered to ....

12. 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?

13. Mathematica gives this partial factorization:

436417 x 2505426087389711 x 94191672740524320175736722723052652107754349558168 26003345520177123345719744592759130647660594274183 378846518589943849834947965570251234101717423

(the last factor is not prime)

14. Originally Posted by janvdl
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?
The prime factors of 104! (only) contain all primes less than 104.
Opalg has already mentioned earlier that 104! + 1 cannot be divisible by any prime less than 105.

But we still do not know what prime factors exist for 104! + 1 larger than 105.

15. Originally Posted by roninpro
Mathematica gives this partial factorization:

436417 x 2505426087389711 x 94191672740524320175736722723052652107754349558168 26003345520177123345719744592759130647660594274183 378846518589943849834947965570251234101717423

(the last factor is not prime)
so,what is this big number?

Page 1 of 2 12 Last