1994 AMC 8 Problem 15

Attempt Problem 15 of the 1994 AMC 8 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 1994 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

If this path is to continue in the same pattern, then which sequence of arrows goes from point 425425 to point 427?427?
Answer: A
Concepts:modular arithmeticpattern recognition
Small hint:
The arrows repeat in a cycle of length 44, so a point’s outgoing arrows depend only on its remainder when divided by 44.
Big hint:
Since 425425 and 427427 leave remainders 11 and 33 when divided by 44, the arrows match those going from point 11 to point 33.
Solution:
The pattern repeats every 44 points, so the arrows leaving a point depend only on its remainder upon division by 4.4.
Because 425=4(106)+1425 = 4(106) + 1 and 427=4(106)+3,427 = 4(106) + 3, the path from 425425 to 427427 looks just like the path from point 11 to point 33: from 11 it goes up to 2,2, then right to 3.3.
Thus, the correct answer is A .
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