Difference between revisions of "2018 AIME I Problems/Problem 11"
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==Solution 6(LTE)== | ==Solution 6(LTE)== | ||
− | We can see that <math>3^n-1 = 143^2*x</math>, which means that <math>v_{11}(3^n-1) \geq 2</math>, <math>v_{13}(3^n-1) \geq 2</math>. <math>v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})</math>, <math>v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})</math> by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as <math>v_{13}(3^n-1) | + | We can see that <math>3^n-1 = 143^2*x</math>, which means that <math>v_{11}(3^n-1) \geq 2</math>, <math>v_{13}(3^n-1) \geq 2</math>. <math>v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})</math>, <math>v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})</math> by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as <math>v_{13}(3^n-1) \geq 2</math>. Therefore the minimum possible value of n is <math>3\times5\times13=\boxed{195}</math>. |
-bradleyguo | -bradleyguo |
Revision as of 16:27, 22 August 2019
Contents
Problem
Find the least positive integer such that when is written in base , its two right-most digits in base are .
Solutions
Modular Arithmetic Solution- Strange (MASS)
Note that the given condition is equivalent to and . Because , the desired condition is equivalent to and .
If , one can see the sequence so .
Now if , it is harder. But we do observe that , therefore for some integer . So our goal is to find the first number such that . In other words, the . It is not difficult to see that the smallest , so ultimately . Therefore, .
The first satisfying both criteria is thus .
-expiLnCalc
Solution 2
Note that Euler's Totient Theorem would not necessarily lead to the smallest and that in this case that is greater than .
We wish to find the least such that . This factors as . Because , we can simply find the least such that and .
Quick inspection yields and . Now we must find the smallest such that . Euler's gives . So is a factor of . This gives . Some more inspection yields is the smallest valid . So and . The least satisfying both is . (RegularHexagon)
Solution 3 (Big Bash)
Listing out the powers of , modulo and modulo , we have:
The powers of repeat in cycles of an in modulo and modulo , respectively. The answer is .
Solution 4(Order+Bash)
We have that Now, so by the Fundamental Theorem of Orders, and with some bashing, we get that it is . Similarly, we get that . Now, which is our desired solution.
Solution 5 (Easy Binomial Theorem)
We wish to find the smallest such that , so we want and . Note that , so repeats with a period of , so . Now, in order for , then . Because , repeats with a period of , so . Hence, we have that for some positive integer , , so and . Thus, we have that , , and , so the smallest possible value of is . -Stormersyle
Solution 6(LTE)
We can see that , which means that , . , by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as . Therefore the minimum possible value of n is .
-bradleyguo
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.