Difference between revisions of "2024 AMC 10A Problems/Problem 23"
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==Solution 4== | ==Solution 4== | ||
− | <math>ab + c = 100 \ | + | <math>ab + c = 100 \circled{ 1}</math> |
− | bc + a = 87 | + | bc + a = 87 \circled{ 2}<math> |
− | ca + b = 60 | + | ca + b = 60 \circled{ 3}</math> |
− | <math>\ | + | <math>\circled{1} + \circled{2} = ab + c +bc + a = (a+c)(b+1)=187</math> |
<math>b+1=\pm 11,\pm 17</math> | <math>b+1=\pm 11,\pm 17</math> | ||
− | <math>\ | + | <math>\circled{1} - \circled{2} = ab + c -bc - a =(a-c)(b-1)=13</math> |
<math>b-1=\pm 1, \pm 13</math> | <math>b-1=\pm 1, \pm 13</math> | ||
Line 68: | Line 68: | ||
<math>\implies a+c=-17</math> | <math>\implies a+c=-17</math> | ||
− | Therefore, <math>ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \ | + | Therefore, <math>ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \circled{1}+\circled{2}+\circled{3} - (a+b+c) = 100+87+60-(a+b+c)</math> |
<math>=\boxed{\textbf{(D) }276}</math> | <math>=\boxed{\textbf{(D) }276}</math> | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] |
Revision as of 10:36, 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
$ab + c = 100 \circled{ 1}$ (Error compiling LaTeX. Unknown error_msg)
bc + a = 87 \circled{ 2}$ca + b = 60 \circled{ 3}$ (Error compiling LaTeX. Unknown error_msg)
$\circled{1} + \circled{2} = ab + c +bc + a = (a+c)(b+1)=187$ (Error compiling LaTeX. Unknown error_msg)
$\circled{1} - \circled{2} = ab + c -bc - a =(a-c)(b-1)=13$ (Error compiling LaTeX. Unknown error_msg)
The only possible pair that has difference of is , , then
Therefore, $ab+ba+ac=ab + c +bc + a + ca + b -(a+b+c) = \circled{1}+\circled{2}+\circled{3} - (a+b+c) = 100+87+60-(a+b+c)$ (Error compiling LaTeX. Unknown error_msg) ~luckuso
~Slightly "latexified" by yuvag
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.