Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"
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Since the power of <math>0</math> to an integer is always <math>0</math>, it | 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 | follows that we want to find the last digit of | ||
− | + | <align*> | |
&2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ | &2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ | ||
&3^2 + 3^{20} + 3^{202} + 3^{2023} | &3^2 + 3^{20} + 3^{202} + 3^{2023} | ||
− | + | </align*> | |
− | |||
Since the powers of <math>2</math> are <math>2, 4, 8, 16, 32</math> | 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 | it follows that <math>2^n</math> and <math>2^{n+4}</math> have the same last | ||
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The expression then has the same last digit as | The expression then has the same last digit as | ||
− | <cmath>^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}</cmath> | + | <cmath>2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}</cmath> |
which is just <math>8</math>. | which is just <math>8</math>. |
Revision as of 12:42, 3 July 2023
Problem
Let and Find the last digit of
Solution
Since the power of to an integer is always , it follows that we want to find the last digit of <align*> &2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ &3^2 + 3^{20} + 3^{202} + 3^{2023} </align*> Since the powers of are it follows that and have the same last digit for . Similarily, and have the same last digit. (This follows as too).
The expression then has the same last digit as which is just .