Just Throw In Some Exponents

Relevant wiki: Diophantine Equations - Solve by Factoring

$\dfrac{a^{3}}{b^{2}}=\dfrac{3}{16}$

$a=\sqrt[3]{\dfrac{3b^{2}}{16}}$

$b=12 \times n^{3}$ where n is a positive integer

$a=3 \times n^{2}$

$\dfrac{b}{a}= 4 \times n$

Since $a<100 , n={{1,2,3,4,5}}$

$\sum\dfrac{b}{a}= 4(1+2+3+4+5) = \boxed{60}$

Step-by-step solution

We have $\frac {a^3}{b^2} = \frac {3}{16} , a<100$ .

Hence, $a^3 = \frac {3}{16}b^2$

Once $a$ is an integer, so must be $a^3$ . Also, $b$ being an integer means that $b^2$ is a natural number. So, in ordem for $a^3$ to be an integer, $b^2$ must be a multiple of 16:

$b^2=16k, k$ natural and a perfect square $(k=z^2, z$ natural)

We can then describe $a^3$ in terms of $k$ :

$a^3 = 3k$

That gives that $k= 9m^3$ , with $m$ being a natural number; so we can take the cubic root of $a^3$ and guarantee that $a$ remains an integer (and also natural, given that it's the cubic root of a natural number, $\frac {3}{16}b^2$ ). Hence:

$a = 3m$ and $b = 12m^{3/2}$

We can now see that $m$ must also be a perfect square, so we can guarantee that $b$ is an integer. Let our final consideration be, then, that $m = n^2, n$ natural. Then, we have, from the first inequality:

$a <100$

$0 \leq 3m < 100$

$0 \leq m < \frac {100}{3}$

As $m$ is natural:

$0 \leq m \leq 33$

$0 \leq n^2 \leq 33$

$0 \leq n \leq \sqrt{33}$

As $n$ is natural:

$0 \leq n \leq 5$

$n \in \{0,1,2,3,4,5\}$

Also, $a=3n^2$ and $b= 12n^3$ . Both terms in which $n=0$ can't be used in our solution, because $\frac {b}{a}$ would be undefined. Finally, we've found 5 pairs of integers forming a sequence $(a_n, b_n) = (3n^2,12n^3), n \in \{1,2,3,4,5\}$ .Therefore:

$\frac {b_n}{a_n} = 4n, n=1,2,3,4,5$

$\displaystyle \sum_{n=1}^5 \frac {b_n}{a_n} = 4\sum_{n=1}^5 n = 60$

The final answer is 60! Hope it helped!

Brute force solution: Each $b_i^2$ can be obtained as $b_i^2=a_i^3\times\frac{16}{3}$ . Notice that $a^3$ must be divisible by 3 for $b^2$ to be an integer. Calculate $b=\sqrt{a^3\times\frac{16}{3}}$ for every $a<100$ multiple of three. Keep the pairs that result in an integer number for the sum $\sum_{i=1}^n \frac{b_i}{a_i}$