Guess Her Age

Let Emily's age be $n = ab$ , where $a$ and $b$ are integers. Then $\dfrac{2}{3} a^2b^2 = m^3$ , where $m$ is an integer. Let $a = 2$ , then $\dfrac{2^3b^2}{3} = m^3$ . It is obvious that $b$ must be a multiple of $3$ and the smallest suitable value is $b=3^2$ , so that we have $\dfrac {2^33^4}{3} = 2^33^3 = 6^3 \quad \Rightarrow n = ab = 2\times 3^2 = \boxed{18}$ , a suitable age for a young lady.

When you multiply by 2/3 you add one to the power of 2 and subtract one from the power of 3. For a number to be a square, it has to have even powers so the smallest number would be 2^2x3^2 (as we know both 2 and 3 are factors). However, a cube has to have powers of multiples of 3. The 2 will become 2^3 when multiplied by 2 but the 3 will become 3^1. Thus, the square must be 2^2x3^4. Therefore her age is 2x3^2 = 2x9 = 18

I got an answer after trial and error:

6^3 = 216.

If Emily's Age= x, then:

(2/3)*(x^2)=216

This implies that x^2=324=18^2.

Hence, she is 18 years old.

I did multiples of 3 and split them into smallest factors. for nine: $\frac{3*3*2}{3}$ I kept doing this until we got three of each factor which are $\frac{3*3*3*3*2*2*2}{3}$