, 2022

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$.