Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"

(Created page with "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 \begin{align*} &2^2 + 2^{20} + 2^{202} + 2^{2...")
 
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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==
 
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*}
 
\begin{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*}
 
\end{align*}
  
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The expression then has the same last digit as
 
The expression then has the same last digit as
\[
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\[^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}
    2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}
 
 
\]
 
\]
 
which is just <math>8</math>.
 
which is just <math>8</math>.

Revision as of 13:40, 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 \begin{align*} &2^2 + 2^{20} + 2^{202} + 2^{2023} + \\ &3^2 + 3^{20} + 3^{202} + 3^{2023} \end{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 \[^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3} \] which is just $8$.