Difference between revisions of "2005 AMC 10A Problems/Problem 14"
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By listing the first two hundred numbers you'll see a pattern. The <math>(-)</math> is where the middle number should go. | By listing the first two hundred numbers you'll see a pattern. The <math>(-)</math> is where the middle number should go. | ||
− | <math>1-0</math> Can't do <math> | + | <math>1-0</math> Can't do <math> \frac{1+0}{2)=\frac{1}{2}</math> |
<math>1-1</math> Can do <math>(1+1)/2=1</math> | <math>1-1</math> Can do <math>(1+1)/2=1</math> |
Revision as of 23:54, 3 January 2016
Problem
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
Solution
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 and slicker)
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
By listing the first two hundred numbers you'll see a pattern. The is where the middle number should go.
Can't do $\frac{1+0}{2)=\frac{1}{2}$ (Error compiling LaTeX. Unknown error_msg)
Can do
Can't do
Can do
Can't do
Can do
Can't do
Can do
Can't do
Can do
Total can do:
Can't do
Can do
Can't do
Can do
Can't do
Can do
Can't do
Can do
Can't do
Can do
Total can do:
SUMMARY
You can now see the pattern. There are can do's each hundred. So then
.
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.