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

## Solutions

### 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 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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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