Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"
Line 5: | Line 5: | ||
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 | ||
− | + | <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} | ||
− | + | </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 | ||
Line 15: | Line 14: | ||
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 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
<align*>
&2^2 + 2^{20} + 2^{202} + 2^{2023} + \
&3^2 + 3^{20} + 3^{202} + 3^{2023}
</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
which is just
.