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
Solution
Since the power of to an integer is always
, it
follows that we want to find the last digit 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
.