Question

Find the unit digit of the numeric expression: ${16}^{23}+{71}^{48}+{59}^{61}$
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\left[2\right]$

Answer 1

The units digit of ${16}^{23}$ is 6, since 6 is in the units digit of the base.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\left[0.5\right]$
The units digit of ${71}^{48}$ is 1, since 1 is in the units digit of the base.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\left[0.5\right]$
The units digit of ${59}^{61}$ is 9, since 9 is in the units digit of the base and the exponent is an odd interger.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\left[0.5\right]$

Sum of all unit places = 6 + 1 + 9 = 16
Units place of the result = 6
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\left[0.5\right]$