Difference between revisions of "2021 Fall AMC 12A Problems/Problem 10"
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==Solution 2 (9's Identity)== | ==Solution 2 (9's Identity)== | ||
− | + | We need to first convert N into a regular base-10 integer: | |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | N&=27{,}006{,}000{,}052_9 \\ | ||
+ | &= 2\cdot9^{10} + 7\cdot9^9 + 6\cdot9^6 + 5\cdot9 + 2 \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Now, consider how the last digit of <math>9</math> changes with changes of the power of <math>9</math>: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 9^0=1 \\ | ||
+ | 9^1=9 \\ | ||
+ | 9^2=81 \\ | ||
+ | 9^3=729 \\ | ||
+ | 9^4=6561 \\ | ||
+ | ...... | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Note that if <math>x</math> is odd: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 9^x &\equiv 4\pmod{5} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | If <math>x</math> is even: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 9^x &\equiv 1\pmod{5} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Therefore, we have: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 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 58 &\pmod{5} \\ | ||
+ | &\equiv \boxed{\textbf{(D) } 3} &\pmod{5} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Hence, we get | ||
~Wilhelm Z | ~Wilhelm Z |
Revision as of 06:26, 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
Recall that We expand by the definition of bases: ~Aidensharp ~kante314 ~MRENTHUSIASM
Solution 2 (9's Identity)
We need to first convert N 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:
Hence, we get
~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.