Problem 507

The prime factorization of $675=3^3\times5^2$, and so 675 has $4\times3=12$ positive factors.
The positive integer $n$ has a
total of $4+14=18$ positive
factors.

Since $n$ has the positive factors
$9=3^2$, 11, $15=3\times5$, and $25=5^2$, then the prime factorization of
$n$ must include at least 2 factors
of 3, at least 2 factors of 5, and at least 1 factor of 11.

In other words, $n$ must be divisible
by $3^2\times5^2\times11$.

Suppose $n=3^2\times5^2\times11$.
Then $n$ has $3\times3\times2=18$ positive factors, as
required.

If $n$ contained additional factors,
then it would have more than 18 positive factors.

Thus, $n=3^2\times5^2\times11=2475$.
Suppose that $m$ is a positive
integer less than 500 that has exactly $2+10=12$ positive factors.

Since $m$ has the positive factors 2
and $9=3^2$, then the prime
factorization of $m$ must include at
least 1 factor of 2, and at least 2 factors of 3.

In other words, $m$ must be divisible
by $2\times3^2$.

To begin, suppose that $m$ has
exactly 2 distinct prime factors.

That is, suppose that $m=2^a\times3^b$ where $a$ and $b$ are integers with $a\geq1$ and $b\geq2$.

In this case, $m$ has $(a+1)(b+1)=12$ positive factors.

Since $a\geq1$ and $b\geq2$, then $a+1\geq2$ and $b+1\geq3$.

Using these restrictions, there are exactly three possibilities for
which $(a+1)(b+1)=12$. These are
$$a+1=2 \text{ and } b+1=6, \text { which gives } a=1 \text{ and } b=5$$ $$a+1=3 \text{ and } b+1=4, \text { which gives } a=2 \text{ and } b=3$$
$$a+1=4 \text{ and } b+1=3, \text { which gives } a=3 \text{ and } b=2$$ If $a=1$ and $b=5$, then $m=2\times3^5=486$.

If $a=2$ and $b=3$, then $m=2^2\times3^3=108$.

If $a=3$ and $b=2$, then $m=2^3\times3^2=72$.

Since each of these values is less than 500, then there are 3 positive
integers that satisfy the given conditions, in this case.

Next, suppose that $m$ has exactly
3 distinct prime factors.

That is, suppose that $\$m=2^a\times3^b\times$
p^c$where$p$ is a prime
number not equal to 2 or 3, and $a$,
$b$ and $c$ are integers with $a\geq1$, $b\geq2$ and $c\geq1$.

If $a=1$, $b=2$ and $c=1$ (the minimum values possible for
$a,b,c$), then $m=2\times3^2\times p$.

In this case, $m$ has $2\times3\times2=12$ positive factors, as
required.

Increasing $a$, $b$ or $c$ increases the number of positive
factors, and thus $a=1$, $b=2$ and $c=1$ is the only possibility for which
$m$ has 12 positive factors and 3
distinct prime factors.

If $a=1$, $b=2$ and $c=1$, then $m=2\times3^2\times p=18p$.

For which prime numbers $p>3$ is
$18p$ less than 500?

Since $18p<500$, then $p<\frac{500}{18}$ and so $p\leq27$.

The prime numbers in this range are 5,7,11,13,17,19, and 23, which give
7 positive integers that satisfy the given conditions, in this case.

Finally, suppose that $m$ has
exactly 4 distinct prime factors.

That is, suppose that $\$m=2^a\times3^b\times p^c\times$ q^d$where$p$ and
$q$ are different prime numbers not
equal to 2 or 3, and $a$, $b$, $c$, $d$
are integers with $a\geq1$, $b\geq2$, $c\geq1$, and $d\geq1$.

If $a=1$, $b=2$, $c=1$, and $d=1$ (the minimum values possible for
$a,b,c,d$), then $m$ has $2\times3\times2\times2=24$ positive
factors, which is a contradiction.

Increasing $a$, $b$, $c$, or $d$ or increasing the number of distinct
prime factors, increases the number of positive factors, and thus there
are no possibilities for which $m$
has 12 positive factors and 4 or more distinct prime factors.

Thus, the number of positive integers less than 500 that have the
factors 2 and 9 and exactly ten other positive factors is $3+7=10$.