Problem concerning the number of 3-digit integers that are both composite and have no prime divisors less than 15

Question

How many three-digit composite integers are not divisible by any prime less than $15$?

I guess this could be written as, if $a$ was a 3-digit integer that is prime factorized as: $a=p_1p_2...p_k$, the prime factors $p_1,p_2,...p_k >15$.

Answer 1

$p_i >15$ so $p_i\ge 17$

$999/17 =<59$ so $p_i <59$ so $p_i\le 57$

$17^3 >999$ so your $k <3$ and as the number is composite $k\ge 2$. So $k=2$

So $a=p_1*p_2$ where $p_1$ and $p_2$ are from $17,19,23,29,31,37,41,43,47,53$

If we assume $p_1\le p_2$ then $p_1 \le p_2 \le \frac {999}{p_1} $.

So if $p_1=17$ there between $17*17$ to $17*53$ there are $10$ acceptable numbers.

$999/19<53$ So between $19*19$ to $19*47$ there are $8$ acceptable numbers.

$999/23\approx 43$ so between $23*23$ to $23*43$ there are $6$ acceptable numbers.

(By the way. $999/23$ is so very close to $43$ that $23*43=989$ is the largest such number.)

$999/29<35$ so between $29*29$ and $29*31$ there are $2$ acceptable numbers.

$999/31\approx 32$ so $31*31$ is an acceptable number. And there can't be any with both factors greater than $31$ as $37*37 >999$.

So there are $27$ possible such composites.

Answer 2

$\sqrt {999}<32$, and $\frac {999}{15}\lt 67$, so $15\lt p_1\le p_2\lt 67$... Consider all products $17\cdot 17,17\cdot 19, 17\cdot23, 17\cdot 29, 17\cdot 31, 17 \cdot 37$ etc formed by primes up to $61$, without both being over $ 32$

(You can save some time by dividing the first prime into $999$, thus getting a bound on the size of the other one )

Sorry I'm too lazy to finish this list right now. .. But repeat what was done for $17$ for the primes up to $61$...