2022 SSMO Speed Round Problems/Problem 1: Difference between revisions
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== | |||
Let <math>S_1 = \{2,0,3\}</math> and <math>S_2 = \{2,20,202,2023\}.</math> Find the last digit of | |||
<cmath>\sum_{a\in S_1,b\in S_2}a^b.</cmath> | |||
==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} + \\ | |||
&3^2 + 3^{20} + 3^{202} + 3^{2023} | |||
\end{align*} | \end{align*} | ||
| Line 11: | Line 15: | ||
The expression then has the same last digit as | 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 <math>8</math>. | which is just <math>8</math>. | ||
Revision as of 12:40, 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*}
&2^2 + 2^{20} + 2^{202} + 2^{2023} + \\
&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
\[^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}
\]
which is just 
.
