Difference between revisions of "2005 AMC 10A Problems/Problem 14"
Sweetmango77 (talk | contribs) (→Solution 2) |
Sweetmango77 (talk | contribs) (→Solution 3) |
||
| Line 40: | Line 40: | ||
==Solution 3== | ==Solution 3== | ||
As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have <math>9</math> possibilities (<math>1-9</math>), and the third digit can have <math>10</math> possibilities (<math>0-9</math>). This means there can be <math>9\cdot10=90</math> possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a <math>0</math> (or any other constant). Exactly half of these would be divisible by <math>2</math>, so our answer is <math>90/2=\textbf{(E) }45</math>. | As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have <math>9</math> possibilities (<math>1-9</math>), and the third digit can have <math>10</math> possibilities (<math>0-9</math>). This means there can be <math>9\cdot10=90</math> possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a <math>0</math> (or any other constant). Exactly half of these would be divisible by <math>2</math>, so our answer is <math>90/2=\textbf{(E) }45</math>. | ||
| + | |||
| + | -[[User:Sweetmango77|SweetMango77]] | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 13:37, 25 November 2020
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
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 
Solution 3
As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have possibilities (
), and the third digit can have 
possibilities (
). This means there can be 
possible three-digit numbers in which the first digit and the third digit are digits and the second digit is a 
(or any other constant). Exactly half of these would be divisible by , so our answer is 
.
Video Solution
CHECK OUT Video Solution: https://youtu.be/eItd9O8cCnQ
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. 
