But Who Ate The Most?

Relevant wiki: Setting Up Equations

The eldest brother took $\dfrac{1}{6}$ of the mangoes, $\dfrac{5}{6}$ are left

The second brother took $\dfrac{1}{5}$ of what the eldest brother left, $\dfrac{5}{6} \times \dfrac{4}{5} = \dfrac{4}{6}$ was left

The third brother took $\dfrac{1}{4}$ of what the second brother left, $\dfrac{4}{6} \times \dfrac{3}{4} = \dfrac{3}{6}$ was left

The fourth brother took $\dfrac{1}{3}$ of what the third brother left, $\dfrac{3}{6} \times \dfrac{2}{3} = \dfrac{2}{6}$ was left

The fifth brother took $\dfrac{1}{2}$ of what the fourth brother left, $\dfrac{2}{6} \times \dfrac{1}{2} = \dfrac{1}{6}$ was left

$3$ mangoes were left. Therefore, $\dfrac{1}{6}$ of the initial number of mangoes is $3$ mangoes.

Let the initial number of mangoes be $x$ . We have

$\dfrac{1}{6} x = 3\\ x=3(6) = \boxed{18}$

Solution strategy: Working backwards

X is the original number of mangoes so at the final stage which is left is X/6 = 3 so X= 18

@Venkatachalam J good problem , it would be more good if you replace the statement How many mangoes were there originally? by How many mangoes were there initially . because the the only possible demanded answer is 18 but by updating the statement you could have a good statement though .