Divide Everything!

i claim that 3 and 5 are the only prime factors. we can check wether it is truth by putting $x\equiv 1,2,0\pmod{3}$ or $x\equiv 1,2,3,4,0\pmod{5}$ and see that it is fulfiled, that case 15 will be the largest divisor. again, as x is even, the expression will be odd implying the exists no even factor. how do we know if there exist no other factors. let the other factor be N where N is an odd integer ≠ 1,3,5,15. then $(x+1)(x+3)(x+5)(x+7)(x+9)\equiv 0 \pmod{N}$ for this to be true over ALL x, it needs to be true for all modulo. so we just put the case $x\equiv 1\pmod{N}$ : $2*4*6*8*10\equiv 2^7*3*5\equiv 0 \pmod{N}$ the modulo implies that N is either even, 1,3 ,5 or15. but it contradicts our original statement. hence the largest divisor is $\boxed{15}$ -contradiction, hence proofed.

Since n is an even number there is a number finishing at 5 between (n+1), (n+3), (n+5), (n+7) and (n+9) $\Rightarrow$ 5 divides its product and 3 also divides $(n+1) \cdot (n+3) \cdot (n+5) \cdot (n+7) \cdot (n+9)$ because 3 divides one only number between (n+1), (n+3) and (n+5) (n is an even number) $\Rightarrow$

gcd ( ( $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ ) ,( $21 \cdot 23 \cdot 25 \cdot 27 \cdot 29$ )) = 15 $\ge$ gcd( $(n+1) \cdot (n+3) \cdot (n+5) \cdot (n+7) \cdot (n+9)$ , for each positive even integer n) $\ge$ lcm(3,5)=15 $\Rightarrow$ 15 = gcd( $(n+1) \cdot (n+3) \cdot (n+5) \cdot (n+7) \cdot (n+9)$ , for each positive even integer n) $\Rightarrow$ 15 is the largest divisor of $(n+1) \cdot (n+3) \cdot (n+5) \cdot (n+7) \cdot (n+9)$ for each positive even integer n.