Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"
(→Solution 2 (9's Identity)) |
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<math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4</math> | <math>\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Modular Arithmetic)== |
Recall that <math>9\equiv-1\pmod{5}.</math> We expand <math>N</math> by the definition of bases: | Recall that <math>9\equiv-1\pmod{5}.</math> We expand <math>N</math> by the definition of bases: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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&= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ | &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ | ||
&\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ | &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 5\cdot(-1) + 2 &&\pmod{5} \\ | ||
− | & | + | &\equiv 2-7+6-5+2 &&\pmod{5} \\ |
− | & | + | &\equiv -2 &&\pmod{5} \\ |
&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. | &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 13:35, 26 November 2021
- The following problem is from both the 2021 Fall AMC 10A #12 and 2021 Fall AMC 12A #10, so both problems redirect to this page.
Problem
The base-nine representation of the number is
What is the remainder when
is divided by
Solution 1 (Modular Arithmetic)
Recall that We expand
by the definition of bases:
~Aidensharp ~kante314 ~MRENTHUSIASM
Solution 2 (9's Identity)
We need to first convert into a regular base-10 integer:
Now, consider how the last digit of changes with changes of the power of
:
Note that if is odd:
If is even:
Therefore, we have:
Note that for the odd case, may simplify the process further, as given by Solution 1.
~Wilhelm Z
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.