Number Theory: Prime Factors A number greater than 5000 but less than 6000 has prime factors $2^x3^y5^z$ such that $x,y,z > 0$. A possible value of $x+y+z$ is:
a)5  b) 8 c)11 d)18
Not sure my approach is works.

 A: We can get the answer from the options given. If we take $x+y+z=5$, then the max possible value is $5^3 ×2^1 ×3^1$ which is 750.
The min possible value for $x+y+z=11$ is $2^9 ×3^1 ×5^1$ is 7680. 
So it can never be 18. The answer is 8.
A: Since we require at least one of each factor $2,3,5$, consider the problem instead as $5000\leq 2\cdot 3\cdot 5\cdot 2^{x'}\cdot 3^{y'}\cdot 5^{z'}\leq 6000$ where we drop the requirement that each $x',y',z'$ be strictly positive and we ask for a possible value of $x'+y'+z'+3$.
We have then $166.\overline{6}\leq 2^{x'}\cdot 3^{y'}\cdot 5^{z'}\leq 200$
Find a lower bound on $x'+y'+z'$ by noting that it will be minimized when $x'=y'=0$ and $z'$ is an appropriate size.  $z'$ will need to be at least $\log_5(166.\overline{6})$, so
$$\log_5(166.\overline{6})\approx 3.1787\leq x'+y'+z'$$
Similarly in finding an upper bound on $x'+y'+z'$ note that it will be maximized when $y'=z'=0$ and $x'$ is an appropriate size.  $x'$ will need to be at most $\log_2(200)$, so
$$x'+y'+z'\leq 7.6438\approx \log_2(200)$$
Thus $6.1787\leq x+y+z=x'+y'+z'+3\leq 10.6438$
Possible values then would be $7,8,9,10$, eliminating options $a,c$ and $d$.
It remains to show that $8$ is still possible, which can be verified via trial and error as $2^33^35^2=5400$
