Question

Which of ${77}^2,{82}^2,{161}^2,{109}^2$ will end with the digit $1$?

Answer 1

Here, the squares of those numbers end with $1$ which end with $1$ or $9$.

Here, ${77}^2=5929,{82}^2=6724,{161}^2=25921,{109}^2=11881.$

Hence, the squares of the numbers ending with the digit $1$ are $161$ and $109$.