Difference between revisions of "2024 AMC 10A Problems/Problem 23"
m |
m |
||
| Line 27: | Line 27: | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}} | ||
| − | {{ | + | {{AMC12 box|year=2024|ab=A|num-b=16|num-a=18}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:42, 8 November 2024
- The following problem is from both the 2024 AMC 10A #23 and 2024 AMC 12A #17, so both problems redirect to this page.
Problem
Integers 
, 
, and 
satisfy 
, 
, and 
. What is 
?
Solution
Subtracting the first two equations yields 
. Notice that both factors are integers, so 
could equal one of 
and 
. We consider each case separately:
For 
, from the second equation, we see that 
. Then 
, which is not possible as is an integer, so this case is invalid.
For 
, we have 
and 
, which by experimentation on the factors of 
has no solution, so this is also invalid.
For 
, we have 
and 
, which by experimentation on the factors of 
has no solution, so this is also invalid.
Thus, we must have 
, so 
and 
. Thus 
, so 
. We can simply trial and error this to find that 
so then 
. The answer is then 
.
~eevee9406
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 | ||
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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. 
