Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"
MRENTHUSIASM (talk | contribs) (→Solution 2 (Powers of 9)) |
m (You don't have to do that here, you can just add all the numbers to get 3.) |
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N&=27{,}006{,}000{,}052_9 \\ | N&=27{,}006{,}000{,}052_9 \\ | ||
&= 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 | + | &\equiv 2\cdot(-1)^{10} + 7\cdot(-1)^9 + 6\cdot(-1)^6 + 2 &&\pmod{5} \\ |
− | &\equiv 2-7+6 | + | &\equiv 2-7+6+2 &&\pmod{5} \\ |
− | |||
&\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. | &\equiv \boxed{\textbf{(D) } 3} &&\pmod{5}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | ~Aidensharp ~Kante314 ~MRENTHUSIASM | + | ~Aidensharp ~Kante314 ~MRENTHUSIASM ~anabel.disher |
==Solution 2 (Powers of 9)== | ==Solution 2 (Powers of 9)== | ||
Line 24: | Line 23: | ||
Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math> | Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9:</math> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 9^0&=1 \\ | + | 9^0&=1, \\ |
− | 9^1&=9 \\ | + | 9^1&=9, \\ |
− | 9^2&= | + | 9^2&=\ldots 1, \\ |
− | 9^3&= | + | 9^3&=\ldots 9, \\ |
− | 9^4&= | + | 9^4&=\ldots 1, \\ |
& \ \vdots | & \ \vdots | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Line 44: | Line 43: | ||
~Wilhelm Z | ~Wilhelm Z | ||
− | ==Video Solution | + | ==Remark== |
− | + | By the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit modulo <math>5</math>, negate it, and add it to the next digit, and repeat the process until you reach the units digit. | |
+ | |||
+ | ~oinava | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/Mv38a8oanFk | ||
~Education, the Study of Everything | ~Education, the Study of Everything |
Latest revision as of 15:38, 13 August 2024
- 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.
Contents
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 ~anabel.disher
Solution 2 (Powers of 9)
We need to first convert into a regular base-
number:
Now, consider how the last digit of changes with changes of the power of
Note that if
is odd, then
On the other hand, if
is even, then
Therefore, we have
Note that for the odd case,
may simplify the process further, as given by Solution 1.
~Wilhelm Z
Remark
By the long division algorithm, you can work from left to right accumulating the answer, and don't need to count the digits to get the power. You only need to take the current leading digit modulo , negate it, and add it to the next digit, and repeat the process until you reach the units digit.
~oinava
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/SCGzEOOICr4?t=101
Video Solution
https://youtu.be/zq3UPu4nwsE?t=358
https://youtu.be/wlDlByKI7A8?t=885
Video Solution by WhyMath
~savannahsolver
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.