Question

How many three-digit positive integers have at least one prime digit?

Answer 1

Answer:

210

Step-by-step explanation:

$512=2^9$ has 9 prime factors

Now, if you want to maximize the number of different prime factor, apply the following method: let $p(n)$  be the nth prime number, in increasing order. Let $v$  define as follow :  $v(0)=1,v(n+1)=v(n)p(n+1)$ , and find n  such that $v(n) < 1000,v(n+1) > 1000$. n will be the number of factors and v(n) your integer.

$v(1)=2v(0)=2v(2)=3v(1)=6v(3)=5v(2)=30v(4)=7v(3)=210v(5)=11v(4)=2310$

There you are,  210 , with only 4 different prime factors.

→Credits: Xavier Dectot←