Difference between revisions of "2024 AMC 10A Problems/Problem 23"
Lord erty09 (talk | contribs) (→Solution) |
Lord erty09 (talk | contribs) (→Solution) |
||
Line 24: | Line 24: | ||
~eevee9406 | ~eevee9406 | ||
− | minor edits by Lord_Erty09 | + | minor edits by Lord_Erty09 |
==Solution 2== | ==Solution 2== |
Revision as of 07:23, 9 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.
Contents
[hide]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
minor edits by Lord_Erty09
Solution 2
Adding up first two equations:
Subtracting equation 1 from equation 2:
Which implies that from
Giving us that
Therefore,
~lptoggled
Solution 3 (Guess and check)
The idea is that you could guess values for , since then and are factors of . The important thing to realize is that , , and are all negative. Then, this can be solved in a few minutes, giving the solution , which gives the answer ~andliu766
Solution 4
ab + c = 100 (1)
bc + a = 87 (2)
ca + b = 60 (3)
(1) + (2) = ab + c +bc + a = (a+c)(b+1)=187
(1) - (2) = ab + c -bc - a =(a-c)(b-1)=13
the only possible pair that has difference of 2 is b-1=-13 , b+1= -11 , then b=-12 , Which implies that Therefore, ~luckuso
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.