2005 AMC 12A Problem 19

Attempt Problem 19 of the 2005 AMC 12A 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 2005 AMC 12A solutions, or check the answer key.

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

19.

A faulty car odometer proceeds from digit 33 to digit 5,5, always skipping the digit 4,4, regardless of position. For example, after traveling one mile the odometer changed from 000039000039 to 000050.000050. If the odometer now reads 002005,002005, how many miles has the car actually traveled?
14041404
14621462
16041604
16051605
18041804
Answer: B
Concepts:number base
Small hint:
The odometer only ever shows 99 distinct digits, so it is really counting in base 99
Big hint:
Map each shown digit to its base-99 value: 5,6,7,8,95, 6, 7, 8, 9 stand for 4,5,6,7,84, 5, 6, 7, 8
Solution:
Because the odometer never displays a 4,4, it uses only 99 symbols and counts in base 9,9, where its digits 5,6,7,8,95, 6, 7, 8, 9 represent the base-99 digits 4,5,6,7,8.4, 5, 6, 7, 8.
The reading 002005002005 therefore corresponds to 20042004 in base 9,9, which equals 2⋅93+4=2⋅729+4=1462. 2 \cdot 9^3 + 4 = 2 \cdot 729 + 4 = 1462.
Thus, the correct answer is B.
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