2013 AIME II Problems/Problem 14
For positive integers and , let be the remainder when is divided by , and for let . Find the remainder when is divided by .
Easy solution without strict proof
We can find that
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
So the sum is , so the answer is .
The solution presented above does not prove why is found by dividing by . Indeed, that is the case, as rigorously shown below.
Consider the case where . We shall prove that . For all , where . This is because and . Also, as n increases, decreases. Thus, for all . Consider all and . Also, . Thus, for for .
Similar proofs apply for and . The reader should feel free to derive these proofs himself.
Highest remainder when is divided by is obtained for and the remainder thus obtained is .
This is the second highest remainder when is divided by and the highest remainder occurs when is divided by = for odd and = for even .
Using the lemma above:
So the answer is
Proof of Lemma: It is similar to stated above.
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