Problem 895

[34]
Since $6000 = 2^4 3^1 5^3$, positive factors of $6000$ can be written as $2^a 3^b 5^c$ where $a = 0, 1, 2, 3, 4$; $b = 0, 1$; $c = 0, 1, 2, 3$. Therefore, there are altogether $5 \cdot 2 \cdot 4 = 40$ positive factors of $6000$. Among them, square numbers arise only when all of $a, b, c$ are even, and this happens only $3 \cdot 1 \cdot 2 = 6$ times. Therefore, there are $40 - 6 = 34$ non-square factors of $6000$.