2007 AMC 12A Problem 16

Question

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

The three digits must form an arithmetic progression in some order

Count the progressions by common difference, then arrange each set, giving fewer arrangements for sets containing

0

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.

Problem 16 in other years