[Number Theory]
Problem: Compute the smallest positive integer $n$ such that $214 \cdot n$ and $2014 \cdot n$ have the same number of divisors.

Solution: Let $D(n)$ be the number of divisors of the integer $n$. Note that if $D(214 n)=D(2014 n)$ and if some $p$ divides $n$ and is relatively prime to both 214 and 2014 , then $D\left(\frac{214 n}{p}\right)=D\left(\frac{2014 n}{p}\right)$. Thus any prime divisor of the smallest possible positive $n$ will be a divisor of $214=2 \cdot 107$ or $2014=2 \cdot 19 \cdot 53$. For the sake of convenience, write $n=2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}$, where $a, b, c, d \geq 1$. Then $D(214 n)=(a+1) b c(d+1)$ and $D(2014 n)=(a+1)(b+1)(c+1) d$. Divide both sides by $a+1$ and expand to get $b c d+b c=b c d+b d+c d+d$, or $b c-b d-c d-d=0$.

Because the goal is to minimize $n$, try $d=1$ : $b c-b-c-1=0 \Rightarrow(b-1)(c-1)=2$, which has solutions $(b, c)=(2,3)$ and $(3,2)$. The latter gives the smaller value for $n$, namely $19^{2} \cdot 53=$ 19133. The only quadruples $(a, b, c, d)$ that satisfy $2^{a-1} 19^{b-1} 53^{c-1} 107^{d-1}<19133$ and $d>1$ are $(1,1,2,2),(1,2,1,2)$, and $(1,1,1,3)$. None of these quadruples satisfies $b c-b d-c d-d=0$, so the minimum value is $n=\mathbf{1 9 1 3 3}$.

Final answer: 19133