# What is the smallest $20$ digit number with at least $13$ distinct prime factors?

The number $10^{19}+1035830$ is a $20$-digit number with $12$ distinct prime factors. I am not sure whether it is the smallest example because to save time I only considered numbers with at least $3$ prime factors below $100$, so I might have overlooked a smaller example.

What is the smallest $20$-digit number with at least $13$ prime factors ?

With brute force, I did not find an example yet. I consider the numbers having at least $5$ prime factors below $100$. Brute force does not seem to be a good way to find the desired number. Does anyone know a more efficient algorithm that guarantees to find the smallest example ?

• $$10^{19}+255252260$$ has $13$ distinct prime factors. Is it the smallest example ? Sep 15, 2017 at 14:45
• By the way, $$10000000118047509570$$ has $14$ distinct prime factors. Is this the smallest example with $20$ digits ? Sep 16, 2017 at 16:41
• And finally, is $$10^{19}+104058800385530$$ the smallest $20$-digit number with $15$ distinct prime factors ? Sep 21, 2017 at 10:45

I got $$10\,000\,000\,000\,255\,252\,260=2^2\cdot 3\cdot5\cdot 7\cdot13\cdot19\cdot37\cdot43\cdot61\cdot73\cdot101\cdot107\cdot1259$$ (with $10$ zeros after the initial $1$!). This number was obtained by multiplying the primes up to $7$, resulting in $210$, and then factoring the multiples of $210$ larger than $10^{19}$ until one was found having $\geq13$ prime factors.
At least one can find some upper bound, i.e., some integer $n\ge 10^{19}$, by multiplying $13$ different primes to an integer $m$, such that $m<10^{19}$, and then using the minimal factor to obtain a number $>10^{19}$. For example, $$m=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\cdot 37\cdot 43=319091739796830$$ gives $$n=31339m=10000016033492855370,$$ having again exactly $13$ different prime divisors. Varying the primes of $m$, one can have better bounds.
• Two questions: (1) Your $m$ seems to have $13$ prime factors...shouldn't it only have $12$. And (2) Is there a reason you used $43$ instead of $41$? Sep 15, 2017 at 15:16
• Yes, $43$ gives a better result than $41$. And also I want that $m$ has already $13$ prime divisors, to make sure, in case the minimal factor does not contain a new prime divisor. Sep 15, 2017 at 15:19