Selection Examinations for the IMO

Write $a_n=n(n+2)(n+4)$ and let $b_n$ be the number of positive divisors of $a_n$. We can easily check that $b_1=4$, $b_2=10$, $b_3=8$, $b_4=14$, $b_5=12$, $b_6=24$, $b_7=12$, $b_8=28$, $b_9=12$ and $b_{10}=40$.

In the next part we will use the following fact: if $p_1^{{\alpha }_1}\cdot p_2^{{\alpha }_2}\cdot \cdots \cdot p_k^{{\alpha }_k}$ is the prime factorization of a positive integer $m$, then the number of positive divisors of $m$ is equal to $$({\alpha }_1+1)({\alpha }_2+1)\dots ({\alpha }_k+1).$$ If $m$ divides a positive integer $l$, then $l$ has at least as many divisors as $m$.

Let $n\ge 11$. If $n$ is even, $n=2k$, then $a_n=2^{3}k(k+1)(k+2)$. At least one of the numbers $k,k+1$ and $k+2$ is divisible by 2 and exactly one of them is divisible by 3. Since $k\ge 6$ the numbers $k,k+1$ and $k+2$ cannot all be powers of 2 or 3. So, $k(k+1)(k+2)$ has a prime divisor $p$ not equal to 2 or 3. Hence, $2^4\cdot 3\cdot p$ divides $a_n$ and this implies that $a_n$ has at least $5\cdot 2\cdot 2=20$ positive divisors.

Let $n\ge 11$ be odd. Then the numbers $n$ and $n+2$ are relatively prime, as are $n+2$ and $n+4$ and also $n$ and $n+4$. One of these three numbers is divisible by 3. This number has at least one other prime divisor $p$ or else is a power of 3. In the latter case it is divisible by $3^3$ since $n\ge 11$. Let $q$ and $r$ be prime divisors of the other two numbers. In the first case the number $a_n$ is divisible by $3pqr$. The numbers $n,n+2$ and $n+4$ are relatively prime, so 3, $p$, $q$ and $r$ are also relatively prime. This implies that $a_n$ has at least $2\cdot 2\cdot 2\cdot 2=16$ divisors. In the second case $a_n$ is divisible by $3^{3}qr$. The primes 3, $q$ and $r$ are again distinct. So, $a_n$ has at least $4\cdot 2\cdot 2=16$ divisors.

The number $a_n$ has at most 15 positive divisors only for $n=1,n=2,n=3,n=4,n=5,n=7$ and $n=9$.