Number of trailing zeros in $50^{3}$ is

Number of trailing zeros in $50^{3}$ is
  1. $2$
  2. $3$
  3. $0$
  4. $6$

Solution

$50^{3} = 125{,}000$. Cubing triples trailing zeros (one zero in $50$ becomes three).

Asked in: IMO

Practice more CUBES AND CUBE ROOTS questions on Aicharya