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.

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11.

A frog begins at P0=(0,0)P_0 = (0, 0) and makes a sequence of jumps according to the following rule: from Pn=(xn,yn),P_n = (x_n, y_n), the frog jumps to Pn+1,P_{n+1}, which may be any of the points (xn+7,yn+2),(x_n + 7, y_n + 2), (xn+2,yn+7),(x_n + 2, y_n + 7), (xn−5,yn−10),(x_n - 5, y_n - 10), or (xn−10,yn−5).(x_n - 10, y_n - 5). There are MM points (x,y)(x, y) with ∣x∣+∣y∣≤100|x| + |y| \le 100 that can be reached by a sequence of such jumps. Find the remainder when MM is divided by 1000.1000.

Answer: 373
Concepts:lattice pointmodular arithmeticparityinvariant
Difficulty rating: 2990
Small Hint:

Each jump changes x+yx + y by +9+9 or −15-15 and changes x−yx - y by +5+5 or −5-5

Big Hint:

Points with x+y=3jx + y = 3j and x−y=5kx - y = 5k need jj and kk of equal parity, and all of them are reachable. Note ∣x∣+∣y∣=max⁡(∣x+y∣,∣x−y∣).|x| + |y| = \max(|x + y|, |x - y|).

Solution:

Each jump changes x+yx + y by +9+9 or −15-15 and changes x−yx - y by ±5.\pm 5. Starting from (0,0),(0, 0), every reachable point therefore has x+y=3jx + y = 3j and x−y=5kx - y = 5k for integers jj and k;k; moreover x=3j+5k2x = \frac{3j + 5k}{2} must be an integer, so jj and kk have the same parity. Since ∣x∣+∣y∣=max⁡(∣x+y∣, ∣x−y∣),|x| + |y| = \max(|x + y|,\, |x - y|), the condition ∣x∣+∣y∣≤100|x| + |y| \le 100 becomes ∣j∣≤33|j| \le 33 and ∣k∣≤20.|k| \le 20.

Conversely, every such point is reachable. Repeating a jump that changes kk in the desired direction first reaches some point on every line x−y=5k.x - y = 5k. Two-jump combinations translate by (9,9)(9, 9) or (−15,−15).(-15, -15). Two of the former plus one of the latter give the shift (3,3),(3, 3), while three of the former plus two of the latter give (−3,−3),(-3, -3), so along a fixed line they move jj by 22 or −2.-2. The initially reached value of jj has the same parity as k,k, so these shifts reach every pair (j,k)(j, k) of equal parity.

Counting: even jj (3333 values) pairs with even kk (2121 values), and odd jj (3434 values) with odd kk (2020 values), so M=33⋅21+34⋅20=1373.M = 33 \cdot 21 + 34 \cdot 20 = 1373. The remainder is 373.373.

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