2018 AIME I Problems/Problem 11
Find the least positive integer such that when is written in base , its two right-most digits in base are .
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 .
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 .
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
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 .
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