Difference between revisions of "2005 AMC 10A Problems/Problem 14"
m (→Solution) |
(→See Also) |
||
Line 38: | Line 38: | ||
The total number, then, is <math>20+25=45\Rightarrow E</math> | The total number, then, is <math>20+25=45\Rightarrow E</math> | ||
− | ==See | + | ==See also== |
− | + | {{AMC10 box|year=2005|ab=A|num-b=13|num-a=15}} | |
− | + | [[Category:Introductory Number Theory Problems]] | |
− | |||
− | |||
− | |||
− | [[Category:Introductory | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:37, 13 August 2019
Problem
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
Solution 1
If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework:
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , possible numbers range from to . So there are numbers in this case.
If the middle digit is , the only possible number is . So there is number in this case.
So the total number of three-digit numbers that satisfy the property is
Solution 2 (much faster, slicker and more awesome)
Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:
If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are numbers in this case.
If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are numbers here.
The total number, then, is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.