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 | ||
| − | + | \begin{align*} | |
| − | &2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ | + | awef&2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ |
| − | &3^2 + 3^{20} + 3^{202} + 3^{2023} | + | awef&3^2 + 3^{20} + 3^{202} + 3^{2023} |
| − | + | \end{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 | ||
Revision as of 12:42, 3 July 2023
Problem
Let 
and 
Find the last digit of
![\[\sum_{a\in S_1,b\in S_2}a^b.\]](http://latex.artofproblemsolving.com/8/b/c/8bc983b7c154bda86c6cf190042b393211940b15.png)
Solution
Since the power of 
to an integer is always , it
follows that we want to find the last digit of
\begin{align*}
awef&2^2 + 2^{20} + 2^{202} + 2^{2023} + \\
awef&3^2 + 3^{20} + 3^{202} + 3^{2023}
\end{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
![\[2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}\]](http://latex.artofproblemsolving.com/b/7/3/b738008166a8f82b045754fa8d3643a26ff14889.png)
which is just 
.
