2007 AMC 12A Problem 16

Below is the professionally curated solution for Problem 16 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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Concepts:arithmetic sequence digits casework

Difficulty rating: 1630

16.

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

$96$

$104$

$112$

$120$

$256$

Solution:

The three distinct digits form an increasing arithmetic progression. Counting by common difference: $8$ with difference $1,$ $6$ with difference $2,$ $4$ with difference $3,$ and $2$ with difference $4,$ for $20$ sets.

Of these, $4$ sets contain $0$ (namely $\{0,1,2\},\{0,2,4\},\{0,3,6\},\{0,4,8\}$ ); each yields $2\cdot 2!=4$ valid numbers since $0$ cannot lead.

The other $16$ sets each yield $3!=6$ numbers. The total is $4\cdot 4+16\cdot 6=112.$

Thus, the correct answer is C.

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