Difference between revisions of "2024 AMC 10B Problems/Problem 24"
(→Solution) |
(→Solution (The simplest way)) |
||
| Line 13: | Line 13: | ||
Their sum is <math>\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1</math>, so <math>P(2023)</math> and <math>P(2024)</math> are also integers. | Their sum is <math>\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1</math>, so <math>P(2023)</math> and <math>P(2024)</math> are also integers. | ||
| − | Therefore, the answer is <math>\boxed{\ | + | Therefore, the answer is <math>\boxed{\textbf{(E) }4}</math>. |
~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx] | ||
Revision as of 06:08, 14 November 2024
Problem
Let
![\[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\]](http://latex.artofproblemsolving.com/2/b/f/2bfe40a41fdb136435709353d32dbe9a87331a66.png)
How many of the values 
, 
, 
, and 
are integers?
Solution (The simplest way)
Here is the English translation with selective LaTeX formatting:
First, we know that and must be integers since they are both divisible by 
. Next, let’s consider the remaining two numbers. Since they are not divisible by , the result of the first term must be a certain number 
, and the result of the second term must be a certain number 
. Similarly, the remaining two terms must each be 
.
Their sum is 
, so and are also integers.
Therefore, the answer is 
.
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
