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