2007 AMC 10B Problem 8

Below is the professionally curated solution for Problem 8 of the 2007 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10B solutions, or check the answer key.

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Concepts:digitsparitycombinations

Difficulty rating: 1290

8.

On the trip home from the meeting where this AMC 10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form bbcac,bbcac, where 0a<b<c9,0\le a\lt b\lt c\le 9, and bb was the average of aa and c.c. How many different five-digit numbers satisfy all these properties?

1212

1616

1818

2020

2424

Solution:

Once aa and cc are chosen, b=a+c2b=\dfrac{a+c}{2} is determined, and a<b<ca\lt b\lt c holds automatically. For bb to be an integer, aa and cc must share parity.

Choosing two even digits from {0,2,4,6,8}\{0,2,4,6,8\} gives (52)=10\binom{5}{2}=10 pairs, and choosing two odd digits from {1,3,5,7,9}\{1,3,5,7,9\} gives another 10.10.

This yields 2020 valid numbers.

Thus, the correct answer is D.

Problem 8 in Other Years