parmenides51 wrote: There are $13$ positive integers greater than $\sqrt{15}$ and less than $\sqrt[3]{B}$. What is the smallest integer value of $B$? Is $a$ an integer? If yes then I'm assuming $a \ge 1$ and $B=16$ doesn't work so $B=17$ and $a=1$ work

@above, there is no $a$, try to increase the browsers' size of letters, it is $\sqrt[3]{B}$, third root of $B$

parmenides51 wrote: @above, there is no $a$, try to increase the browsers' size of letters, it is $\sqrt[3]{B}$, third root of $B$ Oh my bad. Well $3 < \sqrt{15} < 4$ so we want the positive integers to be ${4,5,...,16}$. Therefore $B^{\frac{1}{3}} > 16^3$ so we take $B = 16^3 + 1 = 4097$.

A really easy problem $3<\sqrt{15}<4$, $4,5,6,7,8,9,10,11,12,13,14,15,16<\sqrt[3]{B}\rightarrow B\leq 16^3+$1