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 .
Contents
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 (BigBash)
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.
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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