Selected Problems from Open Contests

Let $N$ be a positive integer that is divisible by $2010$ and that has exactly $2010$ positive divisors. Since $2010=2\cdot 3\cdot 5\cdot 67$, also $N$ should be divisible by these four primes. Thus, $N=2^a\cdot 3^b\cdot 5^c\cdot 67^d\cdot s$, where $a,b,c,d>0$ and $s$ is not divisible by any of the primes $2,3,5,67$. All the factors of $N$ can be expressed as $2^i\cdot 3^j\cdot 5^k\cdot 67^l\cdot t$, where $0\le i\le a,0\le j\le b,0\le k\le c,0\le l\le d$, and $t$ is a factor of $s$. There are $a+1$ choices for $i$ (from $0$ to $a$) and similarly, there are $b+1,c+1$ and $d+1$ choices for $j,k$ and $l$, respectively. Therefore, $N$ has $\delta (N)=(a+1)(b+1)(c+1)(d+1)\delta (s)$ different factors, where $\delta (x)$ stands for the number of factors of $x$. We require $\delta (N)=2010$. As $a+1>1,b+1>1,c+1>1$, and $d+1>1$, we see that each of these numbers is divisible by some prime numbers and the number $\delta (N)=(a+1)(b+1)(c+1)(d+1)\delta (s)$ can thus be expressed as a product of at least four prime numbers. But as $2010$ itself is a product of exactly four prime numbers, we conclude that $a+1,b+1,c+1$, and $d+1$ are exactly those primes $2,3,5$, and $67$, in some order, and $\delta (s)=1$. From the latter condition we see that $s=1$ because any numbers bigger than $1$ has more than one factor. So for $N$ to satisfy the conditions, $N$ must be expressible as $2^a\cdot 3^b\cdot 5^c\cdot 67^d$, where $a,b,c,d$ are the numbers $1,2,4$, and $66$ in some order. Thus there are $4!=24$ numbers satisfying the conditions.