Difference between revisions of "2009 AMC 10B Problems/Problem 21"
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<cmath>\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8</cmath> | <cmath>\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8</cmath> | ||
Therefore, the numerator of the fraction is divisible by <math>8</math>. However, when we divide the numerator by <math>2</math>, we get a remainder of <math>4</math> modulo <math>8</math>, giving us <math>\mathrm{(D)}</math>. | Therefore, the numerator of the fraction is divisible by <math>8</math>. However, when we divide the numerator by <math>2</math>, we get a remainder of <math>4</math> modulo <math>8</math>, giving us <math>\mathrm{(D)}</math>. | ||
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+ | Note: you need to prove that <cmath>\frac{9^{1005}-1}{2}</cmath> is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12 | ||
== See also == | == See also == | ||
{{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:54, 18 January 2021
Problem
What is the remainder when is divided by 8?
Solution
Solution 1
The sum of any four consecutive powers of 3 is divisible by and hence is divisible by 8. Therefore
is divisible by 8. So the required remainder is . The answer is .
Solution 2
We have . Hence for any we have , and then .
Therefore our sum gives the same remainder modulo as . There are terms in the sum, hence there are pairs , and thus the sum is .
Solution 3
We have the formula for the sum of a finite geometric sequence which we want to find the residue modulo 8. Therefore, the numerator of the fraction is divisible by . However, when we divide the numerator by , we get a remainder of modulo , giving us .
Note: you need to prove that is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12
See also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
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