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*}
+
<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}
\end{align*}
+
</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 $S_1 = \{2,0,3\}$ and $S_2 = \{2,20,202,2023\}.$ Find the last digit of \[\sum_{a\in S_1,b\in S_2}a^b.\]

Solution

Since the power of $0$ to an integer is always $0$, 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 $2$ are $2, 4, 8, 16, 32$ it follows that $2^n$ and $2^{n+4}$ have the same last digit for $n \ge 1$. Similarily, $3^n$ and $3^{n+4}$ have the same last digit. (This follows as $\varphi(10) = 4$ too).

The expression then has the same last digit as \[2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}\] which is just $8$.