Difference between revisions of "2005 AMC 12A Problems/Problem 11"

Revision as of 08:40, 1 July 2008

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 $40 + 9 - 4 = 45 \Longrightarrow \mathrm{(E)}$ .

@@ Line 31: / Line 31: @@
 == See also ==
-*[[2007 AMC 12A Problems/Problem 16]]
+*[[2007 AMC 12A Problems/Problem 16|similar problem]]
 {{AMC12 box|year=2005|num-b=10|num-a=12|ab=A}}
 [[Category:Introductory Combinatorics Problems]]

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