Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"
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MRENTHUSIASM (talk | contribs) (→Solution 2 (9's Identity): Reformatted, especially on the align commands.) |
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~Aidensharp ~kante314 ~MRENTHUSIASM | ~Aidensharp ~kante314 ~MRENTHUSIASM | ||
| − | ==Solution 2 (9 | + | ==Solution 2 (Powers of 9)== |
| − | We need to first convert <math>N</math> into a regular base-10 integer: | + | We need to first convert <math>N</math> into a regular base-10 integer: <cmath>N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.</cmath> |
| + | 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 \\ | |
| − | &= 2\ | + | 9^1&=9 \\ |
| + | 9^2&=81 \\ | ||
| + | 9^3&=729 \\ | ||
| + | 9^4&=6561 \\ | ||
| + | & \ \vdots | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | If <math>x</math> is odd, then <math>9^x \equiv 4\pmod{5}.</math> | ||
| − | + | If <math>x</math> is even, then <math>9^x \equiv 1\pmod{5}.</math> | |
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| − | If <math>x</math> is even | ||
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| − | < | ||
| − | 9^x | ||
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| + | Therefore, we have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &\pmod{5} \\ | N&\equiv 2\cdot(1) + 7\cdot(4) + 6\cdot(1) + 5\cdot(4) + 2\cdot(1) &\pmod{5} \\ | ||
&\equiv 2+28+6+20+2 &\pmod{5} \\ | &\equiv 2+28+6+20+2 &\pmod{5} \\ | ||
&\equiv 58 &\pmod{5} \\ | &\equiv 58 &\pmod{5} \\ | ||
| − | &\equiv \boxed{\textbf{(D) } 3} &\pmod{5} \\ | + | &\equiv \boxed{\textbf{(D) } 3} &\pmod{5}. \\ |
\end{align*}</cmath> | \end{align*}</cmath> | ||
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Note that for the odd case, <math>9^x \equiv -1\pmod{5}</math> may simplify the process further, as given by Solution 1. | Note that for the odd case, <math>9^x \equiv -1\pmod{5}</math> may simplify the process further, as given by Solution 1. | ||
Revision as of 13:42, 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 (Powers of 9)
We need to first convert into a regular base-10 integer: ![\[N = 27{,}006{,}000{,}052_9 = 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2.\]](http://latex.artofproblemsolving.com/0/9/9/09964d665dc2054756a6b028ad14afdbe625eff6.png)
Now, consider how the last digit of 
changes with changes of the power of 
If 
is odd, then 
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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
