Difference between revisions of "2024 AMC 10A Problems/Problem 23"
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The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math> | The idea is that you could guess values for <math>c</math>, since then <math>a</math> and <math>b</math> are factors of <math>100 - c</math>. The important thing to realize is that <math>a</math>, <math>b</math>, and <math>c</math> are all negative. Then, this can be solved in a few minutes, giving the solution <math>(-9, -12, -8)</math>, which gives the answer <math>\boxed{\textbf{(D)} 276}</math> | ||
~andliu766 | ~andliu766 | ||
+ | |||
==Solution 4== | ==Solution 4== | ||
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~"latexified" by yuvag | ~"latexified" by yuvag | ||
+ | |||
+ | |||
+ | ==Solution 5== | ||
+ | <math>ab + c = 100 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}</math> | ||
+ | |||
+ | <math>bc + a = 87 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}</math> | ||
+ | |||
+ | <math>ca + b = 60 \text{ } \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}</math> | ||
+ | |||
+ | <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} = ab + c -bc - a =(a-c)(b-1)=13</math> | ||
+ | |||
+ | <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} = bc + a -ca - b =(b-a)(c-1)=27</math> | ||
+ | |||
+ | <math>\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} - \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} = ca + b -ab - c =(c-b)(a-1)=-40</math> | ||
+ | |||
+ | so 3 groups of possible answers | ||
+ | |||
+ | (A) a = 5, b = 14, c = 4 | ||
+ | |||
+ | (B) a = -3, b = -12, c = -3 | ||
+ | |||
+ | (C) a = -9, b = -12 , c = -8 | ||
+ | |||
+ | only group (C) meet all three equation. | ||
+ | |||
+ | Therefore, <math>ab+ba+ac= -9*-12+-12*-8+-8*-9 = \boxed{\textbf{(D) }276}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==See also== | ==See also== |
Revision as of 11:16, 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
The only possible pair that has difference of is , , then
Therefore, ~luckuso
~"latexified" by yuvag
Solution 5
so 3 groups of possible answers
(A) a = 5, b = 14, c = 4
(B) a = -3, b = -12, c = -3
(C) a = -9, b = -12 , c = -8
only group (C) meet all three equation.
Therefore,
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.