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 .
Solutions
Modular Arithmetic Solution- Strange (MASS)
Note that . And . Because , 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 coefficient must be . It is not difficult to see that this first , so ultimately . Therefore, .
The first satisfying both criteria is .
-expiLnCalc
Solution
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)
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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