Difference between revisions of "2005 AMC 12A Problems/Problem 11"
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<math>(\mathrm {A}) \ 41 \qquad (\mathrm {B}) \ 42 \qquad (\mathrm {C})\ 43 \qquad (\mathrm {D}) \ 44 \qquad (\mathrm {E})\ 45</math> | <math>(\mathrm {A}) \ 41 \qquad (\mathrm {B}) \ 42 \qquad (\mathrm {C})\ 43 \qquad (\mathrm {D}) \ 44 \qquad (\mathrm {E})\ 45</math> | ||
| − | + | == Solutions == | |
| − | == | ||
=== Solution 1 === | === Solution 1 === | ||
Let the digits be <math>A, B, C</math> so that <math>B = \frac {A + C}{2}</math>. In order for this to be an integer, <math>A</math> and <math>C</math> have to have the same [[parity]]. There are <math>9</math> possibilities for <math>A</math>, and <math>5</math> for <math>C</math>. <math>B</math> depends on the value of both <math>A</math> and <math>C</math> and is unique for each <math>(A,C)</math>. Thus our answer is <math>9 \cdot 5 \cdot 1 = 45 \implies E</math>. | Let the digits be <math>A, B, C</math> so that <math>B = \frac {A + C}{2}</math>. In order for this to be an integer, <math>A</math> and <math>C</math> have to have the same [[parity]]. There are <math>9</math> possibilities for <math>A</math>, and <math>5</math> for <math>C</math>. <math>B</math> depends on the value of both <math>A</math> and <math>C</math> and is unique for each <math>(A,C)</math>. Thus our answer is <math>9 \cdot 5 \cdot 1 = 45 \implies E</math>. | ||
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This gives us <math>2(8+6+4+2) = 40</math>. However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with <math>0</math>. Thus our answer is <math>40 + 9 - 4 = 45 \Longrightarrow \mathrm{(E)}</math>. | This gives us <math>2(8+6+4+2) = 40</math>. However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with <math>0</math>. Thus our answer is <math>40 + 9 - 4 = 45 \Longrightarrow \mathrm{(E)}</math>. | ||
| − | == See | + | == 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}} | {{AMC12 box|year=2005|num-b=10|num-a=12|ab=A}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:59, 18 October 2020
Problem
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
Solutions
Solution 1
Let the digits be 
so that 
. In order for this to be an integer, 
and 
have to have the same parity. There are 
possibilities for , and 
for . 
depends on the value of both and and is unique for each 
. Thus our answer is 
.
Solution 2
Thus, the three digits form an arithmetic sequence.
- If the numbers are all the same, then there are
possible three-digit numbers. - If the numbers are different, then we count the number of strictly increasing arithmetic sequences between
and 
and multiply by 2 for the decreasing ones:
and multiply by 2 for the decreasing ones:| Common difference | Sequences possible | Number of sequences |
| 1 |  |
8 |
| 2 |  |
6 |
| 3 |  |
4 |
| 4 |  |
2 |
This gives us 
. However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with 
. Thus our answer is 
.
See Also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
