2005 AMC 12A Problems/Problem 11

Problem

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$(\mathrm {A}) \ 41 \qquad (\mathrm {B}) \ 42 \qquad (\mathrm {C})\ 43 \qquad (\mathrm {D}) \ 44 \qquad (\mathrm {E})\ 45$

Solution

Solution 1

Let the digits be $A, B, C$ so that $B = \frac {A + C}{2}$. In order for this to be an integer, $A$ and $C$ have to have the same parity. There are $9$ possibilities for $A$, and $5$ for $C$. $B$ depends on the value of both $A$ and $C$ and is unique for each $(A,C)$. Thus our answer is $9 \cdot 5 \cdot 1 = 45 \implies E$.

Solution 2

Thus, the three digits form an arithmetic sequence.

  • If the numbers are all the same, then there are $9$ possible three-digit numbers.
  • If the numbers are different, then we count the number of strictly increasing arithmetic sequences between $0$ and $10$ and multiply by 2 for the decreasing ones:
Common difference Sequences possible Number of sequences
1 $012, \ldots, 789$ 8
2 $024, \ldots, 579$ 6
3 $036, \ldots, 369$ 4
4 $048, \ldots, 159$ 2

This gives us $2(8+6+4+2) = 40$. However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with $0$. Thus our answer is $40 + 9 - 4 = 45 \Longrightarrow \mathrm{(E)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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