2012 AIME I Problem 11
Attempt Problem 11 of the 2012 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AIME I solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
11.
A frog begins at and makes a sequence of jumps according to the following rule: from the frog jumps to which may be any of the points or There are points with that can be reached by a sequence of such jumps. Find the remainder when is divided by
Answer: 373
Small Hint:
Each jump changes by or and changes by or
Big Hint:
Points with and need and of equal parity, and all of them are reachable. Note
Solution:
Each jump changes by or and changes by Starting from every reachable point therefore has and for integers and moreover must be an integer, so and have the same parity. Since the condition becomes and
Conversely, every such point is reachable. Repeating a jump that changes in the desired direction first reaches some point on every line Two-jump combinations translate by or Two of the former plus one of the latter give the shift while three of the former plus two of the latter give so along a fixed line they move by or The initially reached value of has the same parity as so these shifts reach every pair of equal parity.
Counting: even ( values) pairs with even ( values), and odd ( values) with odd ( values), so The remainder is
Problem 11 in Other Years
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