|
|
| Line 1: |
Line 1: |
| ==Problem==
| |
| Let <math>S_1 = \{2,0,3\}</math> and <math>S_2 = \{2,20,202,2023\}.</math> Find the last digit of
| |
| <cmath>\sum_{a\in S_1,b\in S_2}a^b.</cmath>
| |
| ==Solution==
| |
| Since the power of <math>0</math> to an integer is always <math>0</math>, it
| |
| follows that we want to find the last digit of
| |
| <cmath>2^2 + 2^{20} + 2^{202} + 2^{2023} +3^2 + 3^{20} + 3^{202} + 3^{2023}</cmath>
| |
| Since the powers of <math>2</math> are <math>2, 4, 8, 16, 32</math>
| |
| it follows that <math>2^n</math> and <math>2^{n+4}</math> have the same last
| |
| digit for <math>n \ge 1</math>. Similarily, <math>3^n</math> and <math>3^{n+4}</math> have the same last digit. (This follows as <math>\varphi(10) = 4</math> too).
| |
|
| |
|
| The expression then has the same last digit as
| |
| <cmath>2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}</cmath>
| |
| which is just <math>8</math>.
| |