Question

How many digits (0 to 9) never appear as the last digit of a perfect square?

Answer 1

\(\color{black}{\displaystyle 1\times 1=\color{red}{1} }\)
\(\color{black}{\displaystyle 2\times 2=\color{red}{4} }\)
\(\color{black}{\displaystyle 3\times 3=\color{red}{9} }\)
\(\color{black}{\displaystyle 4\times 4=1\color{red}{6} }\)
\(\color{black}{\displaystyle 5\times 5=2\color{red}{5} }\)
\(\color{black}{\displaystyle 6\times 6=3\color{red}{6} }\)
\(\color{black}{\displaystyle 7\times 7=4\color{red}{9} }\)
\(\color{black}{\displaystyle 8\times 8=6\color{red}{4} }\)
\(\color{black}{\displaystyle 9\times 9=8\color{red}{1} }\)
\(\color{black}{\displaystyle 10\times 10=10\color{red}{0} }\) (don't want to use 0, since 0 is not a perfect sq.)

Answer 2

so it is 3 2 8 7

Answer 3

This way you can determine the last digit of a perfect square.

In other words, (for example)
\(\color{black}{\displaystyle (...\text{_}{\tiny~}\text{_}\color{blue}{4})\times (...\text{_}{\tiny~}\text{_}\color{blue}{4})=(...\text{_}{\tiny~}\text{_}{\tiny~}\text{_}{\tiny~}\text{_}\color{red}{6}) }\)

Answer 4

yes, correct.

Answer 5

so 4

Answer 6

digits

Answer 7

yes, 4 digits:)

Answer 8

thanks

Answer 9

np